[leetcode]Maximum Subarray

Maximum Subarray

 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.

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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：

public class Solution {
    public int maxSubArray(int[] a) {
        if(a == null || a.length == 0) return 0;
        int max = a[0];
        int tem = 0;
        for(int i = 0; i < a.length; i++){
            tem += a[i];
            max = Math.max(tem,max);
            tem = Math.max(tem,0);
        }
        return max;
    }
}

 

思路3：分治。对于每次递归，将数组分半，最大和可能存在

1. 完全在左面部分（递归求解）
2. 完全在右面部分（递归求解）
3. 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：

public class Solution {
     public int maxSubArray(int[] a) {
           if(a == null || a.length == 0) return 0;
           return maxSubArray(a,0,a.length - 1);
       }
       private int maxSubArray(int[] a,int begin ,int end){
           if(begin == end) return a[begin];
           int mid = (begin + end) >> 1;
           int left = maxSubArray(a, begin, mid);
           int right = maxSubArray(a, mid + 1, end);
           int leftHalf = 0,rightHalf = 0;
           int leftMax = Integer.MIN_VALUE,rightMax = Integer.MIN_VALUE;
           for(int i = mid + 1; i <= end; i++){
               rightHalf += a[i];
               rightMax = Math.max(rightHalf, rightMax);
           }
           for(int i = mid; i >= begin; i--){
               leftHalf += a[i];
               leftMax = Math.max(leftMax, leftHalf);
           }
           return Math.max(rightMax + leftMax, Math.max(left, right));
       }
}

分治法也是二分法的一种实现。