【leetcode】Array—— Minimum Size Subarray Sum（209）

题目：

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.

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).

思路 O(n)：维护两个指针，即一个窗口，统计两指针之间数字的累积和。

leetcode解释：Since the given array contains only positive integers, the subarray sum can only increase by including more elements. Therefore, you don't have to include more elements once the current subarray already has a sum large enough. This gives the linear time complexity solution by maintaining a minimum window with a two indices.

代码：

    private int solveN(int s, int[] nums) {        int start = 0, end = 0, sum = 0, minLen = Integer.MAX_VALUE;        while (end < nums.length) {            while (end < nums.length && sum < s) sum += nums[end++];            if (sum < s) break;            while (start < end && sum >= s) sum -= nums[start++];            if (end - start + 1 < minLen) minLen = end - start + 1;        }        return minLen == Integer.MAX_VALUE ? 0 : minLen;    }

思路 N logN：用一个长度为nums.length+1的数组统计nums数组的累计和，该求和数组为递增的，可以用binary search进行查找

leetcode解释：As to NLogN solution, logN immediately reminds you of binary search. In this case, you cannot sort as the current order actually matters. How does one get an ordered array then? 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.

代码：

    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;    }