209 minimum size subarray sum

大于或等于给定值  长度最小的子数组 

leetcode 209

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.

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给定一个整数数组int a[] = {2,3,1,2,4,3},给定一个整数数字s;

求在a中子数组累加和 大于等于s中，长度最小的子数组长度;否则返回0;

一种时间复杂度o(n)的算法。

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思路:两个指针的思路,

start指向目标数组的开始位置,

i指向待查找位置,

[start,i]组成了当前数组已经发现的子数组,子数组可能会随着i一直扩大,

tsum表示子数组的元素和,如果tsum>s说明发现一个解,记下现在子数组的长度[i-start+1]

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code:

class A{
  public:
   int minSubArrayLen(int s, vector<int>& nums) {
        int start = 0;
        int minlen = INT_MAX;
        int tsum = 0;
        for(int i = 0;i<(int)nums.size();i++){
            tsum +=nums[i];
            while(tsum >= s){
                minlen = min(minlen,i-start+1);
                tsum = tsum - nums[start++];///关键是这一步，当tsum>=3是，start下标一直前进，而i下标是不变的
            }
        }
        return minlen==INT_MAX? 0: minlen;
    }
};