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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.
算法思路：
思路1：最土的算法，求出所有的i ~ j的和，继而求出最大值。复杂度O(n^2)
思路2：一维dp，空间可以压缩到O(1)。具体思想看代码：
1 public class Solution { 2 public int maxSubArray(int[] a) { 3 if(a == null || a.length == 0) return 0; 4 int max = a[0]; 5 int tem = 0; 6 for(int i = 0; i < a.length; i++){ 7 tem += a[i]; 8 max = Math.max(tem,max); 9 tem = Math.max(tem,0); 10 } 11 return max; 12 } 13 }
思路3：分治。对于每次递归，将数组分半，最大和可能存在
- 完全在左面部分（递归求解）
- 完全在右面部分（递归求解）
- 横跨左右两个部分（从中间(必须包含中间元素，否则左右无法连接)向两边加，记录最大值）
代码如下：
1 public class Solution { 2 public int maxSubArray(int[] a) { 3 if(a == null || a.length == 0) return 0; 4 return maxSubArray(a,0,a.length - 1); 5 } 6 private int maxSubArray(int[] a,int begin ,int end){ 7 if(begin == end) return a[begin]; 8 int mid = (begin + end) >> 1; 9 int left = maxSubArray(a, begin, mid); 10 int right = maxSubArray(a, mid + 1, end); 11 int leftHalf = 0,rightHalf = 0; 12 int leftMax = Integer.MIN_VALUE,rightMax = Integer.MIN_VALUE; 13 for(int i = mid + 1; i <= end; i++){ 14 rightHalf += a[i]; 15 rightMax = Math.max(rightHalf, rightMax); 16 } 17 for(int i = mid; i >= begin; i--){ 18 leftHalf += a[i]; 19 leftMax = Math.max(leftMax, leftHalf); 20 } 21 return Math.max(rightMax + leftMax, Math.max(left, right)); 22 } 23 }
分治法也是二分法的一种实现。
[leetcode]Maximum Subarray,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/huntfor/p/3903278.html
