# 动态规划 ## 背景 先从一道题目开始~ 如题  [triangle](https://leetcode-cn.com/problems/triangle/) > 给定一个三角形，找出自顶向下的最小路径和。每一步只能移动到下一行中相邻的结点上。 例如，给定三角形： ```text [ [2], [3,4], [6,5,7], [4,1,8,3] ] ``` 自顶向下的最小路径和为  11（即，2 + 3 + 5 + 1 = 11）。 使用 DFS（遍历 或者 分治法） 遍历  分治法  优化 DFS，缓存已经被计算的值（称为：记忆化搜索 本质上：动态规划）  动态规划就是把大问题变成小问题，并解决了小问题重复计算的方法称为动态规划 动态规划和 DFS 区别 - 二叉树 子问题是没有交集，所以大部分二叉树都用递归或者分治法，即 DFS，就可以解决 - 像 triangle 这种是有重复走的情况，**子问题是有交集**，所以可以用动态规划来解决 动态规划，自底向上 ```Python class Solution: def minimumTotal(self, triangle: List[List[int]]) -> int: if len(triangle) == 0: return 0 dp = triangle[-1].copy() for i in range(-2, -len(triangle) - 1, -1): for j in range(len(triangle[i])): dp[j] = triangle[i][j] + min(dp[j], dp[j + 1]) return dp[0] ``` 动态规划，自顶向下 ```Python class Solution: def minimumTotal(self, triangle: List[List[int]]) -> int: if len(triangle) == 0: return 0 dp = triangle[0] for row in triangle[1:]: dp_new = [row[0] + dp[0]] for i in range(len(dp) - 1): dp_new.append(row[i+1] + min(dp[i], dp[i+1])) dp_new.append(row[-1] + dp[-1]) dp = dp_new return min(dp) ``` ## 递归和动规关系 递归是一种程序的实现方式：函数的自我调用 ```go Function(x) { ... Funciton(x-1); ... } ``` 动态规划：是一种解决问题的思想，大规模问题的结果，是由小规模问题的结果运算得来的。动态规划可用递归来实现(Memorization Search) ## 使用场景 满足两个条件 - 满足以下条件之一 - 求最大/最小值（Maximum/Minimum ） - 求是否可行（Yes/No ） - 求可行个数（Count(\*) ） - 满足不能排序或者交换（Can not sort / swap ） 如题：[longest-consecutive-sequence](https://leetcode-cn.com/problems/longest-consecutive-sequence/)  位置可以交换，所以不用动态规划 ## 四点要素 1. **状态 State** - 灵感，创造力，存储小规模问题的结果 2. 方程 Function - 状态之间的联系，怎么通过小的状态，来算大的状态 3. 初始化 Intialization - 最极限的小状态是什么, 起点 4. 答案 Answer - 最大的那个状态是什么，终点 ## 常见四种类型 1. Matrix DP (10%) 1. Sequence (40%) 1. Two Sequences DP (40%) 1. Backpack (10%) > 注意点 > > - 贪心算法大多题目靠背答案，所以如果能用动态规划就尽量用动规，不用贪心算法 ## 1、矩阵类型（10%） ### [minimum-path-sum](https://leetcode-cn.com/problems/minimum-path-sum/) > 给定一个包含非负整数的  *m* x *n*  网格，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。 思路：动态规划 1. state: f(x, y) 从起点走到 (x, y) 的最短路径 2. function: f(x, y) = min(f(x - 1, y), f(x, y - 1]) + A(x, y) 3. intialize: f(0, 0) = A(0, 0)、f(i, 0) = sum(0,0 -> i,0)、 f(0, i) = sum(0,0 -> 0,i) 4. answer: f(n - 1, m - 1) 5. 2D DP -> 1D DP ```Python class Solution: def minPathSum(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) dp = [0] * n dp[0] = grid[0][0] for i in range(1, n): dp[i] = dp[i-1] + grid[0][i] for i in range(1, m): dp[0] += grid[i][0] for j in range(1, n): dp[j] = grid[i][j] + min(dp[j-1], dp[j]) return dp[-1] ``` ### [unique-paths](https://leetcode-cn.com/problems/unique-paths/) > 一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。 > 机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。 > 问总共有多少条不同的路径？ ```Python class Solution: def uniquePaths(self, m: int, n: int) -> int: if m < n: m, n = n, m dp = [1] * n for i in range(1, m): for j in range(1, n): dp[j] += dp[j - 1] return dp[-1] ``` ### [unique-paths-ii](https://leetcode-cn.com/problems/unique-paths-ii/) > 一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。 > 机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。 > 问总共有多少条不同的路径？ > 现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？ ```Python class Solution: def uniquePathsWithObstacles(self, G: List[List[int]]) -> int: m, n = len(G), len(G[0]) dp = [1] if G[0][0] == 0 else [0] for i in range(1, n): new = dp[i-1] if G[0][i] == 0 else 0 dp.append(new) for i in range(1, m): dp[0] = 0 if G[i][0] == 1 else dp[0] for j in range(1, n): dp[j] = dp[j-1] + dp[j] if G[i][j] == 0 else 0 return dp[-1] ``` ## 2、序列类型（40%） ### [climbing-stairs](https://leetcode-cn.com/problems/climbing-stairs/) > 假设你正在爬楼梯。需要  *n*  阶你才能到达楼顶。 ```Python class Solution: def climbStairs(self, n: int) -> int: if n < 2: return n step1, step2 = 2, 1 for _ in range(n - 2): step1, step2 = step1 + step2, step1 return step1 ``` ### [jump-game](https://leetcode-cn.com/problems/jump-game/) > 给定一个非负整数数组，你最初位于数组的第一个位置。 > 数组中的每个元素代表你在该位置可以跳跃的最大长度。 > 判断你是否能够到达最后一个位置。 解法：直接DP无法得到O(n)的解，考虑间接DP - tail to head ```Python class Solution: def canJump(self, nums: List[int]) -> bool: left = len(nums) - 1 # most left index that can reach the last index for i in range(len(nums) - 2, -1, -1): left = i if i + nums[i] >= left else left # DP return left == 0 ``` - head to tail ```Python class Solution: def canJump(self, nums: List[int]) -> bool: max_pos = nums[0] # furthest index can reach for i in range(1, len(nums)): if max_pos < i: return False max_pos = max(max_pos, i + nums[i]) # DP return True ``` ### [jump-game-ii](https://leetcode-cn.com/problems/jump-game-ii/) > 给定一个非负整数数组，你最初位于数组的第一个位置。 > 数组中的每个元素代表你在该位置可以跳跃的最大长度。 > 你的目标是使用最少的跳跃次数到达数组的最后一个位置。 ```Python class Solution: def jump(self, nums: List[int]) -> int: cur_max = 0 step_max = 0 step = 0 for i in range(len(nums)): if cur_max < i: # can't reach i, don't have to consider in this problem return float('inf') if step_max < i: # can't reach i in current number of steps step += 1 step_max = cur_max cur_max = max(cur_max, i + nums[i]) # DP return min_step ``` ### [palindrome-partitioning-ii](https://leetcode-cn.com/problems/palindrome-partitioning-ii/) > 给定一个字符串 _s_，将 _s_ 分割成一些子串，使每个子串都是回文串。 > 返回符合要求的最少分割次数。 - Why is hard 仅目标DP, 判断回文时间复杂度高 -> 目标DP + 回文二维DP, 回文DP空间复杂度高 -> 一点trick, 回文DP空间复杂度降为线性 ```Python class Solution: def minCut(self, s: str) -> int: dp_min = [0] * len(s) dp_pal = [True] * len(s) def isPal(i, j): dp_pal[i] = (s[i] == s[j] and dp_pal[i+1]) return dp_pal[i] for j in range(1, len(s)): min_cut = dp_min[j - 1] + 1 if isPal(0, j): min_cut = 0 for i in range(1, j): if isPal(i, j): min_cut = min(min_cut, dp_min[i - 1] + 1) dp_min[j] = min_cut return dp_min[-1] ``` ### [longest-increasing-subsequence](https://leetcode-cn.com/problems/longest-increasing-subsequence/) > 给定一个无序的整数数组，找到其中最长上升子序列的长度。 - DP(i) 等于以第i个数结尾的最长上升子序列的长度，容易想但不是最优 ```Python class Solution: def lengthOfLIS(self, nums: List[int]) -> int: if len(nums) == 0: return 0 dp_max = [1] * len(nums) for j in range(1, len(nums)): for i in range(j): if nums[j] > nums[i]: dp_max[j] = max(dp_max[j], dp_max[i] + 1) return max(dp_max) ``` - 最优算法使用 greedy + binary search，比较tricky ```Python class Solution: def lengthOfLIS(self, nums: List[int]) -> int: if len(nums) == 0: return 0 seq = [nums[0]] for i in range(1, len(nums)): ins = bisect.bisect_left(seq, nums[i]) if ins == len(seq): seq.append(nums[i]) else: seq[ins] = nums[i] return len(seq) ``` ### [word-break](https://leetcode-cn.com/problems/word-break/) > 给定一个**非空**字符串  *s*  和一个包含**非空**单词列表的字典  *wordDict*，判定  *s*  是否可以被空格拆分为一个或多个在字典中出现的单词。 ```Python class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> bool: dp = [False] * (len(s) + 1) dp[-1] = True for j in range(len(s)): for i in range(j+1): if dp[i - 1] and s[i:j+1] in wordDict: dp[j] = True break return dp[len(s) - 1] ``` 小结 常见处理方式是给 0 位置占位，这样处理问题时一视同仁，初始化则在原来基础上 length+1，返回结果 f[n] - 状态可以为前 i 个 - 初始化 length+1 - 取值 index=i-1 - 返回值：f[n]或者 f[m][n] ## Two Sequences DP（40%） ### [longest-common-subsequence](https://leetcode-cn.com/problems/longest-common-subsequence/) > 给定两个字符串  text1 和  text2，返回这两个字符串的最长公共子序列。 > 一个字符串的   子序列   是指这样一个新的字符串：它是由原字符串在不改变字符的相对顺序的情况下删除某些字符（也可以不删除任何字符）后组成的新字符串。 > 例如，"ace" 是 "abcde" 的子序列，但 "aec" 不是 "abcde" 的子序列。两个字符串的「公共子序列」是这两个字符串所共同拥有的子序列。 - 二维DP若只与当前行和上一行有关，可将空间复杂度降到线性 ```Python class Solution: def longestCommonSubsequence(self, t1: str, t2: str) -> int: if t1 == '' or t2 == '': return 0 if len(t1) < len(t2): t1, t2 = t2, t1 dp = [int(t2[0] == t1[0])] * len(t2) # previous row dp_new = [0] * len(t2) # current row for j in range(1, len(t2)): dp[j] = 1 if t2[j] == t1[0] else dp[j - 1] for i in range(1, len(t1)): dp_new[0] = 1 if dp[0] == 1 or t2[0] == t1[i] else 0 for j in range(1, len(t2)): if t2[j] != t1[i]: dp_new[j] = max(dp[j], dp_new[j - 1]) else: dp_new[j] = dp[j - 1] + 1 dp, dp_new = dp_new, dp return dp[-1] ``` ### [edit-distance](https://leetcode-cn.com/problems/edit-distance/) > 给你两个单词  word1 和  word2，请你计算出将  word1  转换成  word2 所使用的最少操作数   > 你可以对一个单词进行如下三种操作： > 插入一个字符 > 删除一个字符 > 替换一个字符 思路：和上题很类似，相等则不需要操作，否则取删除、插入、替换最小操作次数的值+1 ```Python class Solution: def minDistance(self, w1: str, w2: str) -> int: if w1 == '': return len(w2) if w2 == '': return len(w1) m, n = len(w1), len(w2) if m < n: w1, w2, m, n = w2, w1, n, m dp = [int(w1[0] != w2[0])] * n dp_new = [0] * n for j in range(1, n): dp[j] = dp[j - 1] + int(w2[j] != w1[0] or dp[j - 1] != j) for i in range(1, m): dp_new[0] = dp[0] + int(w2[0] != w1[i] or dp[0] != i) for j in range(1, n): dp_new[j] = min(dp[j - 1] + int(w2[j] != w1[i]), dp[j] + 1, dp_new[j - 1] + 1) dp, dp_new = dp_new, dp return dp[-1] ``` 说明 > 另外一种做法：MAXLEN(a,b)-LCS(a,b) ## 零钱和背包（10%） ### [coin-change](https://leetcode-cn.com/problems/coin-change/) > 给定不同面额的硬币 coins 和一个总金额 amount。编写一个函数来计算可以凑成总金额所需的最少的硬币个数。如果没有任何一种硬币组合能组成总金额，返回  -1。 思路：和其他 DP 不太一样，i 表示钱或者容量 ```Python class Solution: def coinChange(self, coins: List[int], amount: int) -> int: dp = [0] * (amount + 1) for i in range(1, len(dp)): dp[i] = float('inf') for coin in coins: if i >= coin and dp[i - coin] + 1 < dp[i]: dp[i] = dp[i - coin] + 1 return -1 if dp[amount] == float('inf') else dp[amount] ``` ### [backpack](https://www.lintcode.com/problem/backpack/description) > 在 n 个物品中挑选若干物品装入背包，最多能装多满？假设背包的大小为 m，每个物品的大小为 A[i] ```Python class Solution: def backPack(self, m, A): n = len(A) dp = [0] * (m + 1) dp_new = [0] * (m + 1) for i in range(n): for j in range(1, m + 1): use_Ai = 0 if j - A[i] < 0 else dp[j - A[i]] + A[i] dp_new[j] = max(dp[j], use_Ai) dp, dp_new = dp_new, dp return dp[-1] ``` ### [backpack-ii](https://www.lintcode.com/problem/backpack-ii/description) > 有 `n` 个物品和一个大小为 `m` 的背包. 给定数组 `A` 表示每个物品的大小和数组 `V` 表示每个物品的价值. > 问最多能装入背包的总价值是多大? 思路：dp(i, j) 为前 i 个物品，装入 j 背包的最大价值 ```Python class Solution: def backPackII(self, m, A, V): n = len(A) dp = [0] * (m + 1) dp_new = [0] * (m + 1) for i in range(n): for j in range(1, m + 1): use_Ai = 0 if j - A[i] < 0 else dp[j - A[i]] + V[i] # previous problem is a special case of this problem that V(i) = A(i) dp_new[j] = max(dp[j], use_Ai) dp, dp_new = dp_new, dp return dp[-1] ``` ## 补充 ### [maximum-product-subarray](https://leetcode-cn.com/problems/maximum-product-subarray/) > 最大乘积子串 处理负数情况稍微有点复杂，注意需要同时 DP 正数乘积和负数乘积 ```Python class Solution: def maxProduct(self, nums: List[int]) -> int: max_product = float('-inf') dp_pos, dp_neg = 0, 0 for num in nums: if num > 0: dp_pos, dp_neg = max(num, num * dp_pos), dp_neg * num else: dp_pos, dp_neg = dp_neg * num, min(num, dp_pos * num) if dp_pos != 0: max_product = max(max_product, dp_pos) elif dp_neg != 0: max_product = max(max_product, dp_neg) else: max_product = max(max_product, 0) return max_product ``` ### [decode-ways](https://leetcode-cn.com/problems/decode-ways/) > 1 到 26 分别对应 a 到 z，给定输入数字串，问总共有多少种译码方法 常规 DP 题，注意处理edge case即可 ```Python class Solution: def numDecodings(self, s: str) -> int: def valid_2(i): if i < 1: return 0 num = int(s[i-1:i+1]) return int(num > 9 and num < 27) dp_1, dp_2 = 1, 0 for i in range(len(s)): dp_1, dp_2 = dp_1 * int(s[i] != '0') + dp_2 * valid_2(i), dp_1 return dp_1 ``` ### [best-time-to-buy-and-sell-stock-with-cooldown](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/) > 给定股票每天的价格，每天可以买入卖出，买入后必须卖出才可以进行下一次购买，卖出后一天不可以购买，问可以获得的最大利润 经典的维特比译码类问题，找到状态空间和状态转移关系即可 ```Python class Solution: def maxProfit(self, prices: List[int]) -> int: buy, buy_then_nothing, sell, sell_then_nothing = float('-inf'), float('-inf'), float('-inf'), 0 for p in prices: buy, buy_then_nothing, sell, sell_then_nothing = sell_then_nothing - p, max(buy, buy_then_nothing), max(buy, buy_then_nothing) + p, max(sell, sell_then_nothing) return max(buy, buy_then_nothing, sell, sell_then_nothing) ``` ### [word-break-ii](https://leetcode-cn.com/problems/word-break-ii/) > 给定字符串和可选的单词列表，求字符串所有的分割方式 思路：此题 DP 解法容易想但并不是好做法，因为和 word-break 不同，此题需要返回所有可行分割而不是找到一组就可以。这里使用 个人推荐 backtrack with memoization。 ```Python class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> List[str]: n = len(s) result = [] mem = collections.defaultdict(list) wordDict = set(wordDict) def backtrack(first=0, route=[]): if first == n: result.append(' '.join(route)) return True if first not in mem: for next_first in range(first + 1, n + 1): if s[first:next_first] in wordDict: route.append(s[first:next_first]) if backtrack(next_first, route): mem[first].append(next_first) route.pop() if len(mem[first]) > 0: return True elif len(mem[first]) > 0: for next_first in mem[first]: route.append(s[first:next_first]) backtrack(next_first) route.pop() return True return False backtrack() return result ``` ### [burst-balloons](https://leetcode-cn.com/problems/burst-balloons/) > n 个气球排成一行，每个气球上有一个分数，每次戳爆一个气球得分为该气球分数和相邻两气球分数的乘积，求最大得分 此题主要难点是构造 DP 的状态，过程为逆着气球戳爆的顺序 ```Python class Solution: def maxCoins(self, nums: List[int]) -> int: n = len(nums) nums.append(1) dp = [[0] * (n + 1) for _ in range(n + 1)] for dist in range(2, n + 2): for left in range(-1, n - dist + 1): right = left + dist max_coin = float('-inf') left_right = nums[left] * nums[right] for j in range(left + 1, right): max_coin = max(max_coin, left_right * nums[j] + dp[left][j] + dp[j][right]) dp[left][right] = max_coin nums.pop() return dp[-1][n] ``` ## 练习 Matrix DP (10%) - [ ] [triangle](https://leetcode-cn.com/problems/triangle/) - [ ] [minimum-path-sum](https://leetcode-cn.com/problems/minimum-path-sum/) - [ ] [unique-paths](https://leetcode-cn.com/problems/unique-paths/) - [ ] [unique-paths-ii](https://leetcode-cn.com/problems/unique-paths-ii/) Sequence (40%) - [ ] [climbing-stairs](https://leetcode-cn.com/problems/climbing-stairs/) - [ ] [jump-game](https://leetcode-cn.com/problems/jump-game/) - [ ] [jump-game-ii](https://leetcode-cn.com/problems/jump-game-ii/) - [ ] [palindrome-partitioning-ii](https://leetcode-cn.com/problems/palindrome-partitioning-ii/) - [ ] [longest-increasing-subsequence](https://leetcode-cn.com/problems/longest-increasing-subsequence/) - [ ] [word-break](https://leetcode-cn.com/problems/word-break/) Two Sequences DP (40%) - [ ] [longest-common-subsequence](https://leetcode-cn.com/problems/longest-common-subsequence/) - [ ] [edit-distance](https://leetcode-cn.com/problems/edit-distance/) Backpack & Coin Change (10%) - [ ] [coin-change](https://leetcode-cn.com/problems/coin-change/) - [ ] [backpack](https://www.lintcode.com/problem/backpack/description) - [ ] [backpack-ii](https://www.lintcode.com/problem/backpack-ii/description)