Maximum Subarray

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#leetcode #最大子段和 #算法

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本文介绍了一种寻找数组中具有最大和的连续子数组的方法，通过动态规划算法实现，给出两个C++代码示例。

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题目

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.

思路

数组的最大子段和问题（连续的一段子数组）

动态规划：

令b[j] = MAX { Sum(A[i]+...+A[j]) } 1<=j<=n

1<=i<=j

所以如果b[j-1]>0 , b[j] = b[j-1]+A[j] ; 否则 b[j] = A[j] 。

代码一

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];
        int b = A[0];
        for(int i=1;i<n;i++)
        {
            if(b>0)
                b += A[i];
            else
                b = A[i];
            if(b>sum)
                sum = b;            
        }
        return sum;    
        
    }
};

代码二

对于b[j] = max { max{ b[j-1]+A[j] , A[j] } , A[j] }

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];
        int b = A[0];
        for(int i=1;i<n;i++)
        {
            b = max(b+A[i],A[i]);
            b = max(b,A[i]);            
            if(b>sum)
                sum = b;            
        }
        return sum;    
        
    }
};