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

.

click to show more practice.

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.

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。

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;
}
}

Reference:

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