Leetcode Maximum Subarray

经典最大子串和问题，要求用两种方法实现：1、Kadane算法，复杂度O(n)； 2、分治法，复杂度O(nlogn)。

1、Kadane算法：代码比较简单，但是理解需要时间。初始化ans为0， 每次用ans加上A[i]的值，并更新最大值，如果遇到ans<0，更新ans的值为0。

class Solution {
public:
    int maxSubArray(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int ans = 0, maxn = INT_MIN;
        for(int i = 0; i < n; i++){
            if(ans < 0) ans = 0;
            ans += A[i];
            maxn = max(maxn, ans); 
        }
        return maxn;
    }
};

2、分治法：最大子串和的区间有以下三种情况（low，high分别为左右边界，mid为(low+high)/2）：

(1) 区间完全在 A[low,mid-1]
(2) 区间完全在 A[mid+1,high]
(3) 区间包含有 A[mid]

按照这三种情况一直递归下去即可。

class Solution {
public:
    int divide(int A[], int low, int high){
        if(low == high) return A[low];
        if(low == high-1)
            return max(A[low]+A[high], max(A[low], A[high]));
        int mid = (low+high)/2;
        int lmax = divide(A, low, mid-1);
        int rmax = divide(A, mid+1, high);
        int mmax = A[mid];
        int tmp = mmax;
        for(int i = mid-1; i >=low; i--){
            tmp += A[i];
            if(tmp > mmax)  mmax = tmp;
        }
        tmp = mmax;
        for(int i = mid+1; i <= high; i++){
            tmp += A[i];
            if(tmp > mmax)  mmax = tmp;
        }
        return max(mmax, max(lmax, rmax));
        
    }
    int maxSubArray(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        return divide(A, 0, n-1);
    }
    
};