[LeetCode49]Maximum Subarray

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

Analysis:

O(n)就是一维DP.
假设A(0, i)区间存在k，使得[k, i]区间是以i结尾区间的最大值， 定义为Max[i], 在这里，当求取Max[i+1]时，
Max[i+1] = Max[i] + A[i+1],  if (Max[i] + A[i+1] >0)
                = 0, if(Max[i]+A[i+1] <0)，如果和小于零，A[i+1]必为负数，没必要保留，舍弃掉
然后从左往右扫描，求取Max数字的最大值即为所求。

Java

public int maxSubArray(int[] A) {int max = 0;int res = Integer.MIN_VALUE;for(int i=0;i<A.length;i++){max = max>=0 ? (max+A[i]):A[i];res = Math.max(max, res);}return res;}

int maxSubArray(int A[], int n) {int sum = 0;int res = INT_MIN;for(int i=0;i<n;i++){sum = sum>=0?(sum+A[i]):A[i];res = max(sum,res);}return res;}

采用Divide & Conquer。这就暗示了，解法必然是二分。分析如下：

假设数组A[left, right]存在最大值区间[i, j](i>=left & j<=right)，以mid = (left + right)/2 分界，无非以下三种情况：

subarray A[i,..j] is
(1) Entirely in A[low,mid-1]
(2) Entirely in A[mid+1,high]
(3) Across mid
对于(1) and (2)，直接递归求解即可，对于(3)，则需要以min为中心，向左及向右扫描求最大值，意味着在A[left, Mid]区间中找出A[i..mid], 而在A[mid+1, right]中找出A[mid+1..j]，两者加和即为(3)的解。

Java

public int maxSubArray(int[] A) {int res  = Integer.MIN_VALUE;return getMaxSubarray(A, 0, A.length-1, res);}public int getMaxSubarray(int A[],int left, int right, int smax){if(left>right)return Integer.MIN_VALUE;int mid = (left+right)/2;int lmax = getMaxSubarray(A,left,mid-1,smax);int rmax = getMaxSubarray(A,mid+1,right,smax);smax = Math.max(lmax,smax);smax = Math.max(rmax,smax);int sum =0, mlmax = 0;for(int i=mid-1;i>=left;i--){sum+=A[i];if(sum>mlmax)mlmax = sum;}sum = 0;int mrmax = 0;for(int i=mid+1;i<=right;i++){sum+=A[i];if(sum>mrmax)mrmax = sum;}smax = Math.max(smax, mlmax+mrmax+A[mid]);return smax;}

int getMaxSubarray(int A[],int left, int right, int smax){if(left>right)return INT_MIN;int mid = (left+right)/2;int lmax = getMaxSubarray(A,left,mid-1,smax);int rmax = getMaxSubarray(A,mid+1,right,smax);smax = max(lmax,smax);smax = max(rmax,smax);int sum =0, mlmax = 0;for(int i=mid-1;i>=left;i--){sum+=A[i];if(sum>mlmax)mlmax = sum;}sum = 0;int mrmax = 0;for(int i=mid+1;i<=right;i++){sum+=A[i];if(sum>mrmax)mrmax = sum;}smax = max(smax, mlmax+mrmax+A[mid]);return smax;
}int maxSubArray(int A[], int n) {int maxv = INT_MIN;return getMaxSubarray(A,0,n-1,maxv);}