子数组最大累加和+子矩阵的最大和+子数组累加和为给定值的最大子数组长度

1.如何求一个子数组的最大累加和？（时间复杂度O(N)）

class Solution{public:int SubArrayMaxSum(int *a, size_t n){if ((a == NULL) || (n == 0))return 0;int max = a[0];int cur = 0;for (size_t i = 0; i < n; i++){cur += a[i];max = (max>cur ? max : cur);  //当max > cur时，需要更新maxif (cur < 0)cur = 0;                 //当cur < 0 时，需要将cur置0。}return max;}};

2.如何求两个子数组的最大累加和？(要求两个子数组无重合的部分,时间复杂度O(N))

class Solution{public:int SubArrayMaxSum(int *a,  size_t n){if ((a == NULL) || (n < 1))return 0;int maxR = a[n - 1];int curR = a[n - 1];vector<int> maxSubSumR;maxSubSumR.push_back(maxR);int maxSubSum = a[0];int cur = a[0];for (int i = n - 2; i >= 0; i--){curR += a[i];maxR = (maxR > curR ? maxR : curR);  //当max > cur时，需要更新maxmaxSubSumR.push_back(maxR);if (curR < 0)curR = 0;                 //当cur < 0 时，需要将cur置0。}int max = a[0] + maxSubSumR[n - 2];int j = 1;for (int i = n - 3; i >= 0; i--){cur += a[j];maxSubSum = (maxSubSum>cur ? maxSubSum : cur);max = (max > (maxSubSum + maxSubSumR[i]) ? max : (maxSubSum + maxSubSumR[i]));j++;if (cur < 0)cur = 0;}return max;}};

3.未排序的正数数组中累加和为给定值的最长子数组长度（时间复杂度O(N)）

class   Solution {public:int SubArrayMaxLegth(int* a, int n, int key){if (a == NULL || n < 0 || key < 0)return -1;int left = -1;int length = 0;int maxLength = 0;int  right = -1;int sum = 0;//循环结束的条件left==right&& left==n-1||right > n-1while ((left != right) || (left != n - 1)){if (sum > key){left++;sum -= a[left];length--;}else{if (sum == key){maxLength = maxLength > length ? maxLength : length;}right++;length++;if (right > n - 1)break;sum += a[right];}}return maxLength;}};

4.子矩阵的最大和问题求解

#define  M 100#define  N 100class Solution{public:int SubMatrixMax(int a[4][4], int m, int n){if (a == NULL)return 0;int SubMatrixValue[M][N] = { 0 };//SubMatrixValue数组的大小可以根据m*n二维数组的大小来确定for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (i == 0)SubMatrixValue[i][j] = a[i][j];elseSubMatrixValue[i][j] = SubMatrixValue[i - 1][j] + a[i][j];}}//求出了1——>m行每一列的子数组中每一列的和int max = SubMatrixValue[0][0];int cur = SubMatrixValue[0][0];for (int k = -1; k < m; k++){for (int i = k + 1; i < m; i++){if (k != -1){for (int j = 0; j < n; j++){SubMatrixValue[i][j] -= SubMatrixValue[k][j];}}cur = SubMatrixValue[i][0];for (int j = 0; j < n; j++){if (i != 0)cur += SubMatrixValue[i][j];elsecur = SubMatrixValue[i][0];if (cur > max)max = cur;if (cur < 0)cur = 0;}}}return max;}};