分治求解最大子数组

分治求解步骤

1. 分：将数组分成左右两部分
2. 治：递归地求解左半部分和右半部分的最大子数组
3. 合：计算跨越中点的最大子数组，并取三者中的最大值

具体实现

1. 分：

   将数组A分成两部分

   左半部分：从low到mid

   右半部分：从mid+1 到high

   mid为low+high的中点

2. 治

   递归求解左半部分和右半部分的最大子数组

3. 合

   计算跨越中点的最大子数组。这需要从中点县向左和向右分别扩展，找到包括中点的最大子数组

#include <limits.h>
#include <stdio.h>int findMaxCrossingSubarrary(int A[], int low, int mid, int high, int *maxLeft, int * maxRight)
{int leftSum = INT_MIN;int sum = 0;int i;for(i = mid; i >= low; i--){sum += A[i];if(sum > leftSum){leftSum = sun;*maxLeft = i;}}
}int findMaxSubarrary(int A[], int low, int high, int *maxLeft, int *maxRight)
{if(low == high){*maxLeft = low;*maxRight = high;return A[low];}else{int mid = (low + high) / 2;int leftLow, leftHigh, rightLow, rightHigh, crossLow, crossHigh;int leftSum = findMaxSubarrary(A, low, mid, &leftLow, &leftHigh);int rightSum = findMaxSubarrary(A, mid + 1, high, &rightLow, &rightHigh);int crossSum = findMaxCrossingSubarrary(A, low, mid, high, &crossLow, &crossHigh);if(leftSum >= rightSum && leftSum >= crossSum){*maxLeft = leftLow;*maxRight = leftHigh;return leftSum;}else if(rightSum >= leftSum && rightSum >= crossSum){*maxLeft = rightLow;*maxRight = rightHigh;return rightSum;}else{*maxLeft = crossLow;*maxRight = crossHigh;return crossSum;}}
}int main()
{int A[] = {13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7};int n = sizeof(A) / sizeof(A[0]);int low = 0;int high = n - 1;int maxLeft, maxRight;int maxSum = findMaxSubarrary(A, low, high, &maxLeft, &maxRight);printf("maximum subarry: [%d, %d] with sum %d\n", maxLeft, maxRight, maxSum);return 0;
}

分：在 findMaxSubarray 函数中，通过计算 mid 将数组分成左右两部分。

治：递归调用 findMaxSubarray 函数分别求解左半部分和右半部分的最大子数组。

合：调用 findMaxCrossingSubarray 函数计算跨越中点的最大子数组。将左右两部分和跨越中点的最大子数组中的最大值作为结果返回