【LeetCode】Maximum Subarray_leetcode最大子串和-优快云博客

参考链接

http://blog.youkuaiyun.com/magisu/article/details/14515209

题目描述

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.

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.

题目分析

题目是意思是，找出一个子串，要求子串中数字之和最大。

这是一个经典的动态规划的题目，需要了解以下规律。

max_subarray(A[n]) = max( max_subarray(A[n-1]), A[n], max_subarray(A[n-1)(+)A[n] );

最大子串和是之前已经遍历过的数字中的最大值max_subarray(A[n-1])，当前值A[n]，和加上当前值max_subarray(A[n-1)(+)A[n]这三者之间的最大值。

代码示例

#include 
   
    
using namespace std;

#if 1
class Solution {  
public:  
    int maxSubArray(int A[], int n) {  
        // Start typing your C/C++ solution below  
        // DO NOT write int main() function  
        int sum = A[0], max = A[0];  
        for (int i = 1; i < n; i++) {  
            sum = sum < 0 ? A[i] : sum + A[i];  
            max = sum > max ? sum : max;  
        }  
        return max;  
    }  
};  
#elif 1//自己写的，数据大时就会出问题 
class Solution {
public:
    int maxSubArray(int A[], int n) {
        int *sum = new int[n];
        for(int i = 0;i
    
     0;j--)
        	{
	        	if(i>=j)
	        	{
	        		sum[j]=sum[j-1]+A[i];
	        		if(sum[j] > max_sum) max_sum = sum[j];
	        	}
        	}
        	sum[0] = A[i];
        	if(sum[0] > max_sum) max_sum = sum[0];
        	/*
        	printf("%4d:",A[i]);
        	for(int k = 0;k