分治策略 | 最大子数组问题

分治策略  就是把一个大问题递归地分解成规模较小的子问题，子问题的规模可能不等。

最大子数组问题，找数组A中和最大的连续段A[i, .., j]。A中要有负数这个问题才有意义。

伪代码：

（跨中点的情况）

 

（完全在原数组左边或右边的情况）

  

C++代码：

#include <iostream>using namespace std;struct maxSubarray {int low;int high;int sum;};maxSubarray findMaxCrossingSubarray(int* arr, int low, int mid, int high){int sum = 0;int leftSum = -65535;int rightSum = -65535;maxSubarray crossSub;crossSub.low = low;crossSub.high = high;for (int i = mid; i >= low; i--){sum += arr[i];if (sum > leftSum){leftSum = sum;crossSub.low = i;}}sum = 0;for (int i = mid + 1; i <= high; i++){sum += arr[i];if (sum > rightSum){rightSum = sum;crossSub.high = i;}}crossSub.sum = leftSum + rightSum;return crossSub;}maxSubarray findMaxmumSubarray(int* arr, int low, int high, int sum){if (low == high){maxSubarray sub;sub.low = low;sub.high = high;sub.sum = arr[low];return sub;}else{int mid = (low + high) / 2;maxSubarray leftSub;leftSub = findMaxmumSubarray(arr, low, mid, sum);maxSubarray rightSub;rightSub = findMaxmumSubarray(arr, mid + 1, high, sum);maxSubarray crossSub;crossSub = findMaxCrossingSubarray(arr, low, mid, high);if (leftSub.sum >= rightSub.sum && leftSub.sum >= crossSub.sum)return leftSub;else if (rightSub.sum >= leftSub.sum && rightSub.sum >= crossSub.sum)return rightSub;else if (crossSub.sum >= leftSub.sum && crossSub.sum >= rightSub.sum)return crossSub;}}int main(){int A[10] = { 4, 5, 7, -2, -3, -6, -1, 9, -2, 4 };maxSubarray maxSub;maxSub = findMaxmumSubarray(A, 0, sizeof(A) / sizeof(A[0]) - 1, -65535);cout << maxSub.low << "," << maxSub.high << "," << maxSub.sum << endl;return 0;}

---------- 更新分治策略写法 (上面的代码是何等的woc...) ----------

int maxSubArray(vector<int>& nums) {    if (nums.empty())        return 0;    return findMaxSubArray(nums, 0, nums.size() - 1);}int findMaxSubArray(vector<int>& nums, int left, int right) {    if (left >= right) return nums[left];    int mid = left + (right - left) / 2;    int lmax = findMaxSubArray(nums, left, mid - 1);    int rmax = findMaxSubArray(nums, mid + 1, right);    int mmax = nums[mid], temp = mmax;               //跨中点数组    for (int i = mid - 1; i >= left; --i) {        temp += nums[i];        mmax = max(mmax, temp);    }    temp = mmax;    for (int i = mid + 1; i <= right; ++i) {        temp += nums[i];        mmax = max(mmax, temp);    }    return max(mmax, max(lmax, rmax));}

习题中也介绍了一种方法，线性时间O(n)指的是已知A[1,..,j]的最大子数组再求A[1,..,j+1]的最大子数组，但如果看整体的时间复杂度，还是O(n^2)，不如分治法好。