Minimum Size Subarray Sum -- leetcode

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.

基本思路：滑动窗口，双指针

一个指针在前，进行不断的累加；

当累加和超过指定阀值后，一个指针在后，不断累减，直到累加和小于给定阀值。

形象的看起来，有如一个保持阀值宽度的窗口，从左向右滑动。

O(n)

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

Minimum Size Subarray Sum -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/47355857