209. Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous 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.

题意：给一个数组和值s,求数组里连续子序列和>=s,并使得这个连续子序列长度最小。

思想：由于是连续的子序列 可以定义2个指针start,end，tempsum为数组[start....end]之和的值,min为所有tempsum>=s的最小值

  移动end或start指针

  若数组[start....end]之和<s 那么指针end++;

  若数组[start....end]之和>s 判断min是否是最小值 然后指针start++; 

  找到tempsum>=s时的最小值min

C++ AC代码：Time O(n) Space O(1)   这次有些判断不到位，导致代码复杂度比较高，还需加强训练！！！

class Solution {public:    int minSubArrayLen(int s, vector<int>& nums) {        int len = nums.size();        if(len<=0) return 0;        int min = len;        int b = 0;        int e = 0;        int tempsum = nums[0];        while(b<=e&&e<len){            if(e==b&&nums[e]>=s)                return 1;            else if(e==b&&nums[e]<s){                    e++;                    tempsum +=nums[e];            }            else if(e!=b&&tempsum>=s){                min = min<(e-b+1)?min:(e-b+1);if(e==len-1&&tempsum-nums[b]<s)break;                tempsum-=nums[b];                b++;            }else if(e!=b&&tempsum<s){                e++;if(e==len)break;                tempsum +=nums[e];            }        }        min = (tempsum>=s||min!=len)?min:0;        return min;    }};