Maximum Subarray [LEETCODE]

最新推荐文章于 2019-08-02 13:24:00 发布

转载最新推荐文章于 2019-08-02 13:24:00 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/scenix/p/3369912.html

本文介绍如何使用O(n)复杂度找到数组中连续子数组的最大和，包括实例演示和不同解题思路。

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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.

====================================================================================

Consider an array A[i,....,j,j+1]. If we know its lagest subarrya is A[i,...,j].

Then the new array's lagest sum subarray may same as the array A[i,...j], or is A[i,...j,j+1].

That's only depends on if A[j+1] is larger than sum(A[i,...j]).

 1 class Solution {
 2 public:
 3     int maxSubArray(int A[], int n) {
 4         // Note: The Solution object is instantiated only once and is reused by each test case.
 5        int max_sum = INT_MIN;
 6        int sum = 0;
 7        for(int i = 0; i < n; i++) {
 8            sum = max(sum + A[i], A[i]);
 9            max_sum = max(max_sum, sum);
10        }
11        return max_sum;
12 
13     }
14 };