LeetCode 209. Minimum Size Subarray Sum（最小子数组之和）

原题网址：https://leetcode.com/problems/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.

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More practice:

If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

方法一：用二分法尝试各个长度的子数组是否满足条件。

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int i=1, j=nums.length;
        boolean found = false;
        while (i<=j) {
            int m = (i+j)/2;
            if (window(s, nums, m)) { 
                j = m-1; 
                found = true; 
            } else {
                i=m+1;
            }
        }
        return found ? i : 0;
    }
    private boolean window(int s, int[] nums, int w) {
        int sum = 0;
        for(int i=0; i<nums.length; i++) {
            if (i>=w) sum -= nums[i-w];
            sum += nums[i];
            if (sum>=s) return true;
        }
        return false;
    }
}

方法二：使用移动窗口。

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int min = 0;
        int sum = 0;
        int from = 0;
        for(int i=0; i<nums.length; i++) {
            sum += nums[i];
            while (sum >= s && from <= i) {
                if (min == 0 || i-from+1<min) min = i-from+1;
                sum -= nums[from++];
            }
        }
        return min;
    }
}