Given an array ofnpositive integers and a positive integers, find the minimal length of acontiguoussubarray of which the sum ≥s. If there isn't one, return 0 instead.

For example, given the array[2,3,1,2,4,3]ands = 7,
the subarray[4,3]has the minimal length under the problem constraint.

Solution 1: j右移直到sum > = target，然后右移i，每次更新min，直到sum < target。

    public int minSubArrayLen(int s, int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
        for (;j < nums.length; j++) {
            sum += nums[j];
            //这个if可省去
            if (sum < s) {
                continue;
            }
            while (sum >= s) {
                min = Math.min(min, j - i + 1);
                sum -= nums[i];
                i++;
            }
        }
        if (min == Integer.MAX_VALUE) {
            return 0;
        }
        return min;
    }

Solution 2: 先求sum[]，Since all elements are positive, the cumulative sum must be strictly increasing. Then, a subarray sum can expressed as the difference between two cumulative sum. Hence, given a start index for the cumulative sum array, the other end index can be searched using binary search. NlogN

    private int solveNLogN(int s, int[] nums) {
        int[] sums = new int[nums.length + 1];
        for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i - 1];
        int minLen = Integer.MAX_VALUE;
        for (int i = 0; i < sums.length; i++) {
            int end = binarySearch(i + 1, sums.length - 1, sums[i] + s, sums);
            if (end == sums.length) break;
            if (end - i < minLen) minLen = end - i;
        }
        return minLen == Integer.MAX_VALUE ? 0 : minLen;
    }

    private int binarySearch(int lo, int hi, int key, int[] sums) {
        while (lo <= hi) {
           int mid = (lo + hi) / 2;
           if (sums[mid] >= key){
               hi = mid - 1;
           } else {
               lo = mid + 1;
           }
        }
        return lo;
    }

Solution 3: O(NLogN) - search if a window of size k exists that satisfy the condition

    public int minSubArrayLen(int s, int[] nums) {
        int i = 1, j = nums.length, min = 0;
        while (i <= j) {
            int mid = (i + j) / 2;
            if (windowExist(mid, nums, s)) {
                j = mid - 1;
                min = mid;
            } else i = mid + 1;
        }
        return min;
    }


    private boolean windowExist(int size, int[] nums, int s) {
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            if (i >= size) sum -= nums[i - size];
            sum += nums[i];
            if (sum >= s) return true;
        }
        return false;
    }