Leetcode's Medium collection (C + + implementation) eight

1 Pow (x, N)
The problem is recursive by dichotomy

Double Mypow (doublex,intN) {if(n==0)return 1;if(n<0) {n= (-N);x=1/x; } Double Res=mypow (x, n/2);if(n%2==0)        {returnRes*res; }Else{returnRes*res*x; }    }

2 Maximum Subarray
Find the contiguous subarray within an array (containing at least one number) which have the largest sum. For example, given the array [? 2,1,?3,4,?1,2,1,?5,4], the contiguous subarray [4,?1,2,1] have the largest sum = 6.
The problem can be solved by dynamic programming, d P[I]=d P[I?1]>0?d P[I?1]+Nums[I]:Nums[I] . Obviously, when the current maximum value is less than 0 o'clock, the next value is calculated less than 0 of the number discarded.

int maxSubArray(vector<int>& nums) {        int max=INT_MIN;        int n=nums.size();        int mid=0;        for(int i=0;i<n;i++)        {        //这里用mid,max替代dp数组以节省空间            mid=mid+nums[i];            mid=(nums[i]>=mid?nums[i]:mid);            if(mid>max)            {                max=mid;            }        }        return max;    }

Leetcode's Medium collection (C + + implementation) eight