【leetcode】Maximum Subarray

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.

 

在累加的过程中，如果发现sum<0则说明前面的序列对后面的序列没有贡献，故此时设置sum=0

 

class Solution {
public:
    int maxSubArray(int A[], int n) {
       
        int sum,maxSum;
        sum=maxSum=A[0];
        for(int i=1;i<n;i++)
        {
            if(sum<0) sum=0;
            sum+=A[i];
           
            if(maxSum<sum) maxSum=sum;
        }
        return maxSum;       
    }
};

 

采用分治法求解：

找到左半边最大的序列值，找到右半边最大的序列值，找到中间序列的值

 

class Solution {
public:
    int maxSubArray(int A[], int n) {
       
        divideAndConquer(A,0,n-1);
    }
   
    int divideAndConquer(int A[],int left,int right)
    {
       
        if(left==right) return A[left];
 
        int mid=(left+right)/2;
       
       
        int leftMax=divideAndConquer(A,left,mid);
        int rightMax=divideAndConquer(A,mid+1,right);
       
        int midSum1=0;
        int midMax1=A[mid];
       
        for(int i=mid;i>=left;i--)
        {
            midSum1+=A[i];
            if(midMax1<midSum1) midMax1=midSum1;
        }
       
        int midSum2=0;
        int midMax2=A[mid+1];
       
        for(int i=mid+1;i<=right;i++)
        {
            midSum2+=A[i];
            if(midMax2<midSum2) midMax2=midSum2;
        }
       
        int midMax=midMax1+midMax2;
       
        return max(max(leftMax,rightMax),midMax);
       
       
    }
};