[Array]Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

For example, given the array [2,3,1,2,4,3] and s = 7,
the subarray [4,3] has the minimal length under the problem constraint.

方法1：利用TwoPointer的思想进行遍历，时间复杂度为O(n)。

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int start = 0,sum = 0,minlength=INT_MAX;
        for(int i=0;i<nums.size();++i){
            sum+=nums[i];
            while(sum>=s){
                minlength = min(minlength,i-start+1);
                sum-=nums[start++];
            }
        }
        return minlength==INT_MAX?0:minlength;
    }
};

方法2：因为数组中的元素都是正数，那么对于数组中每一个位置上的元素，记录该位置前所有元素的累加和，并将这些累加和保存到一个vector中去，利用这个累加和vector进行操作，可以发现这个vector中的元素是递增性质，可以利用这个性质进行二分查找，这里我们可以发现accumulate[j]-accumulate[i]为i+1到j之间的累加和。正好满足subarray这个概念。这种方法的时间复杂度为O（nlog(n)）。

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        vector<int> accumulate(nums.size()+1,0);
        int minLen = INT_MAX;
        int sum = 0;
        for(int i=1;i<accumulate.size();++i)
            accumulate[i]=accumulate[i-1]+nums[i-1];
        for(int i=0;i<accumulate.size()-1;++i){
            int target = s + accumulate[i];
            int pos = upperbound(accumulate,target,i+1,accumulate.size()-1);
            if(pos==-1)
                continue;
            minLen = min(minLen,pos-i);
        }
        return minLen==INT_MAX?0:minLen;
    }
private:
    int upperbound(vector<int>& accumulate,int target,int left,int right){
        if(accumulate[right]<target)
            return -1;
        while(left<right){
            int middle = (left+right)/2;
            if(accumulate[middle]<target)
                left=middle+1;
            else
                right=middle;
        }
        return left;
    }
};