【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[];
         for(int i=;i<n;i++)
         {
             if(sum<) sum=;
             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,,n-);
     }
 
     int divideAndConquer(int A[],int left,int right)
     {
 
         if(left==right) return A[left];
 
         int mid=(left+right)/;
 
         int leftMax=divideAndConquer(A,left,mid);
         int rightMax=divideAndConquer(A,mid+,right);
 
         int midSum1=;
         int midMax1=A[mid];
 
         for(int i=mid;i>=left;i--)
         {
             midSum1+=A[i];
             if(midMax1<midSum1) midMax1=midSum1;
         }
 
         int midSum2=;
         int midMax2=A[mid+];
 
         for(int i=mid+;i<=right;i++)
         {
             midSum2+=A[i];
             if(midMax2<midSum2) midMax2=midSum2;
         }
 
         int midMax=midMax1+midMax2;
 
         return max(max(leftMax,rightMax),midMax);
 
     }
 };