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/*
53. Maximum Subarray
Easy
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.
Example:
Input: [-2,1,-3,4,-1,2,1,-5,4],
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
Follow up:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
*/
/*
分治解题思路：
1. 二分数组
2. 找出左边最大和
3. 找出右边最大和
4. 找出包含mid的最大值
5. 输出三者中的最大值
*/
package dc;
public class MaximumSubarray {
public static int maxSubArray(int[] nums) {
return subMax(nums, 0, nums.length - 1);
}
private static int subMax(int[] nums, int left, int right) {
if (left >= right) {
return nums[left];
}
int mid = left + (right - left) / 2;
// 二分左区间最大和
int maxLeft = 0;
// 二分右区间最大和
int maxRight = 0;
// 包含中间值，向左右延伸子串最大和
int maxLeftMid = 0;
maxLeft = subMax(nums, 0, mid - 1);
maxRight = subMax(nums, mid + 1, right);
maxLeftMid = midMax(nums, left, right);
return Math.max(Math.max(maxLeft, maxRight), maxLeftMid);
}
private static int midMax(int[] nums, int left, int right) {
if (left >= right) {
return nums[left];
}
int mid = left + (right - left) / 2;
int max = nums[mid];
int preMax = max;
for (int i = mid - 1; i >= left; i--) {
preMax += nums[i];
if (max < preMax) {
max = preMax;
}
}
int rightMax = nums[mid + 1];
preMax = rightMax;
for (int i = mid + 2; i <= right; i++) {
preMax += nums[i];
if (rightMax < preMax) {
rightMax = preMax;
}
}
if (rightMax > 0) {
max += rightMax;
}
return max;
}
}