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/*
@filename 8.cpp
@author caonan
@date 2022-03-18 07:43:10
@reference 剑指offer专项
@url https://leetcode-cn.com/problems/2VG8Kg/
@brief 给定一个含有 n 个正整数的数组和一个正整数 target,
找出该数组中满足其和 ≥ target 的长度最小的 连续子数组 [numsl, numsl+1, ..., numsr-1, numsr] ，
并返回其长度。如果不存在符合条件的子数组，返回 0 。
*/
#include <assert.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
using namespace std;
class Solution
{
public:
//代码还是不够简洁
int minSubArrayLen(int target, vector<int>& nums)
{
int i = 0;
int j = i;
int sum = 0;
int min = 0x80000000 - 1;
while (i <= j && j < nums.size()) {
sum += nums[j];
while (sum >= target && i <= j) {
min = std::min(j - i + 1, min);
sum -= nums[i++];
}
j++;
}
return min == 0x80000000 - 1 ? 0 : min;
}
//改用for循环更加简洁一点点
int minSubArrayLen1(int target, vector<int>& nums)
{
int j = 0;
int sum = 0;
int min = 0x80000000 - 1;
for (int i = 0; i < nums.size(); i++) {
sum += nums[i];
while (sum >= target) {
min = std::min(i - j + 1, min);
sum -= nums[j++];
}
}
return min == 0x80000000 - 1 ? 0 : min;
}
};
int main()
{
Solution s;
vector<int> nums{2, 3, 1, 2, 4, 3};
assert(s.minSubArrayLen(7, nums) == 2);
assert(s.minSubArrayLen1(7, nums) == 2);
vector<int> nums1{1, 4, 4};
assert(s.minSubArrayLen(4, nums1) == 1);
assert(s.minSubArrayLen1(4, nums1) == 1);
vector<int> nums2{1, 1, 1, 1, 1, 1, 1, 1};
assert(s.minSubArrayLen(11, nums2) == 0);
assert(s.minSubArrayLen1(11, nums2) == 0);
return 0;
}