[Leetcode] maximun subarray 最大子数组

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.

题意：给定由正负整数组成的数组，问和最大的子数组，返回和。

思路：这题和jump game、jump game ii 类似。定义两个变量，一个维护目前的最大值maxValue，一个是之前元素值的最大值sum，当sum小于0时说明，若算上之前的数会是整体的最大值减小，所以舍弃。时间复杂度为O(n)。代码如下：

class Solution {
public:
    int maxSubArray(int A[], int n) 
    {
        if(n==0)    return 0;
        int sum=0,maxValue=A[0];
        for(int i=0;i<n;++i)
        {
            sum+=A[i];
            maxValue=max(maxValue,sum);
            if(sum<0)   
                sum=0;
        }    
        return maxValue;
    }
};

方法二：分治。参考Grandyang的博客。分治的思想类似二分搜索，首先将数组一分为二，分别找出左边和右边的最大子数组之和，然后从中间开始向左右分别扫描，求出的最大值分别和左右两边得到的最大值相比较，取最大值。

代码如下：

class Solution {
public:
    int maxSubArray(int A[], int n) 
    {
        if(n==0)    return 0;
        return helper(A,0,n-1);
    }

    int helper(int A[],int left,int right)
    {
        if(left>=right) return A[left];
        int mid=(left+right)>>1;
        int lmx=helper(A,left,mid-1);
        int rmx=helper(A,mid+1,right);

        int mMax=A[mid],temp=mMax;

        //遍历左半端，
        for(int i=mid-1;i>=left;--i)
        {
            temp+=A[i];
            mMax=max(mMax,temp);
        }
        temp=mMax;  //向右
        for(int i=mid+1;i<=right;++i)
        {
            temp+=A[i];
            mMax=max(mMax,temp);
        }

        return max(mMax,max(lmx,rmx));
    }
};