分治法解决最大子序和
今天初尝分支法,试了以下分治法求最大子序和,感觉确实比动态规划好理解一点点
int getMaxSum2(int a[], int n) {
if (n == 1) {
return a[0];
}
return get(a, 0, n - 1);
}
int get(int a[], int left, int right) {
//如果leftt等于right,说明只有这一个元素,返回它
if (left == right) {
return a[left];
}
//递归得求两边得最大和,每次求和都分两半
int mid = (left + right) / 2;
int leftSum = get(a, left, mid);
int rightSum = get(a, mid + 1, right);
//所以每一个子数组的最大子序列在左边子数组,右边子数组,横跨中间的三种情况中出现
int MaxLeftSum = 0;
int MaxRightSum = 0;
int tempMidSum = 0;
for (int i = mid; i >= left; i--) {
tempMidSum += a[i];
if (tempMidSum > MaxLeftSum) {
MaxLeftSum = tempMidSum;
}
}
tempMidSum = 0;
for (int i = mid+1; i <= left; i++) {
tempMidSum += a[i];
if (tempMidSum > MaxRightSum) {
MaxRightSum = tempMidSum;
}
}
int MaxMidSum = MaxRightSum + MaxLeftSum;
return max(max(leftSum,rightSum),MaxMidSum);
}