Maximum Subarray leetcode java

题目：

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

More practice:

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

题解：

这道题要求求连续的数组值，加和最大。

试想一下，如果我们从头遍历这个数组。对于数组中的其中一个元素，它只有两个选择：

1. 要么加入之前的数组加和之中（跟别人一组）

2. 要么自己单立一个数组（自己单开一组）

所以对于这个元素应该如何选择，就看他能对哪个组的贡献大。如果跟别人一组，能让总加和变大，还是跟别人一组好了；如果自己起个头一组，自己的值比之前加和的值还要大，那么还是自己单开一组好了。

所以利用一个sum数组，记录每一轮sum的最大值，sum[i]表示当前这个元素是跟之前数组加和一组还是自己单立一组好，然后维护一个全局最大值即位答案。

代码如下；

1     public int maxSubArray(int[] A) {
2         int[] sum = new int[A.length];
3
4         int max = A[0];
5         sum[0] = A[0];
6
7         for (int i = 1; i < A.length; i++) {
8             sum[i] = Math.max(A[i], sum[i - 1] + A[i]);
9             max = Math.max(max, sum[i]);
10         }
11
12         return max;
13     }

同时发现，这道题是经典的问题，是1977布朗的一个教授提出来的。

http://en.wikipedia.org/wiki/Maximum_subarray_problem

并发现，这道题有两种经典解法，一个是：Kadane算法，算法复杂度O(n)；另外一个是分治法：算法复杂度为O(nlogn)。

1. Kadane算法

代码如下：

1     public int maxSubArray(int[] A) {
2         int max_ending_here = 0;
3         int max_so_far = Integer.MIN_VALUE;
4
5         for(int i = 0; i < A.length; i++){
6             if(max_ending_here < 0)
7                  max_ending_here = 0;
8             max_ending_here += A[i];
9             max_so_far = Math.max(max_so_far, max_ending_here);
10         }
11         return max_so_far;
12     }

2. 分治法：

代码如下：

1     public int maxSubArray(int[] A) {
2          return divide(A, 0, A.length-1);
3     }
4
5   public int divide(int A[], int low, int high){
6         if(low == high)
7             return A[low];
8         if(low == high-1)
9             return Math.max(A[low]+A[high], Math.max(A[low], A[high]));
10
11         int mid = (low+high)/2;
12         int lmax = divide(A, low, mid-1);
13         int rmax = divide(A, mid+1, high);
14
15         int mmax = A[mid];
16         int tmp = mmax;
17         for(int i = mid-1; i >=low; i--){
18             tmp += A[i];
19             if(tmp > mmax)
20                 mmax = tmp;
21         }
22         tmp = mmax;
23         for(int i = mid+1; i <= high; i++){
24             tmp += A[i];
25             if(tmp > mmax)
26                 mmax = tmp;
27         }
28         return Math.max(mmax, Math.max(lmax, rmax));
29
30     }

Reference：

http://en.wikipedia.org/wiki/Maximum_subarray_problem

http://www.cnblogs.com/statical/articles/3054483.html

http://blog.csdn.net/xshengh/article/details/12708291

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