Leetcode 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.

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解题思路：

Dynamic Programming.

sum[i] = max(nums[i], nums[i] + sum[i-1])

 

The changing condition for dynamic programming is "We should ignore the sum of the previous n-1 elements if nth element is greater than the sum."

 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。

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Java code:

第一种写法：

public class Solution {
    public int maxSubArray(int[] nums) {
        int len = nums.length;
        int[] sum = new int[len];
        int max = nums[0];
        sum[0] = nums[0];
        for(int i = 1; i< len; i++) {
            sum[i] = Math.max(nums[i], nums[i] + sum[i-1]);
            max = Math.max(max, sum[i]);
        }
        return max;
    }
}

第二种写法： sum不用数组，只用一个int

public class Solution {
    public int maxSubArray(int[] nums) {
        int sum = nums[0];
        int max = nums[0];
        for(int i = 1; i< nums.length; i++) {
            sum = Math.max(nums[i], nums[i] + sum);
            max = Math.max(max, sum);
        }
        return max;
    }
}

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Reference:

1. http://www.programcreek.com/2013/02/leetcode-maximum-subarray-java/

2. http://www.programcreek.com/2013/02/leetcode-maximum-subarray-java/