晨哥Leetcode 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.
Example:
Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.
Follow up:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

解析：
subarray sum的问题肯定先想到prefix sum
此题是找最小的长度，相当于最小的窗口，很自然想到滑动窗口法
指针i， j， 右边的j 一直向右滑动，直到窗口之和 > target为止
满足后左边的i再开始向右收缩这个窗口

注意边界条件！j == a.length, 已经走到底了，但是循环还没结束，左边i还需要继续向右滑动.
如果sum<s, 那左边收缩就没意义了，可以退出

public int minSubArrayLen(int s, int[] a) {
        int sum = 0, min = Integer.MAX_VALUE;
        int i = 0, j=0;
        for(;i<a.length;i++) {
            while(j<a.length && sum < s) {
                sum+=a[j++];
            }
            if(sum >= s) {
                min = Math.min(min, j-i);
            } else{
            	break; // 已经到底都满足不了，可以放弃了，再收缩没意义
            }
            //if(j == a.length) break; 右边到底了，左边还可以继续收缩呢，别退出
            sum -= a[i];
        }
        return min == Integer.MAX_VALUE ? 0 : min;
    }

follow up, 要求O(n logn)算法
那肯定想到要排序，或者binary search
因为数组里的数都是正数，和是递增的，其实用Binary search就可以
比较懒，写了个treemap的，复杂度一样

public int minSubArrayLen(int s, int[] a) {
        int sum = 0, min = Integer.MAX_VALUE;
        TreeMap<Integer, Integer> map = new TreeMap();
        map.put(0, -1);
        for(int i = 0;i<a.length;i++) {
            sum+=a[i];
            map.put(sum, i);
        }
        for(Map.Entry<Integer, Integer> entry: map.entrySet()) {
            Map.Entry<Integer, Integer> ceiling = map.ceilingEntry(entry.getKey() + s);
            if(ceiling != null) {
                min = Math.min(min, ceiling.getValue() - entry.getValue());
            } else {
                break;
            }
        }
        return min == Integer.MAX_VALUE ? 0 : min;
    }