209. Minimum Size Subarray Sum (M)

Minimum Size Subarray Sum (M)

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).

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题意

给定一个正整数s和一个正整数数组nums，要求在nums中找到一个最短的连续子序列，使其和大于等于s。

思路

**滑动窗口法(Two Pointers)：**连续子数组问题很容易想到使用滑动窗口来解题。用left和right分别指向窗口的左端点和右端点，如果当前窗口内的整数和小于s，则将右端点右移来拉长窗口；如果当前窗口内的整数和大于等于s，判断其长度是否比当前结果小，是则记录，并将左端点右移来缩短窗口。重复以上步骤就能得到一个长度最短且整数和大于等于s的窗口。最坏情况下左右端点都走了一遍数组，所以时间复杂度为$O(2N)=O(N)$。

**二分查找：**新建sum数组，$sum[i]=nums[0]+...+nums[i]$，则问题转化为在数组sum中求一个连续子数组，使其末首之差大于等于s。从左向右遍历，每次固定左端点i，在右半边子数组中利用二分法找到值大于等于$s+sum[i]$的元素sum[j]，判断长度并记录。时间复杂度为$O(NlogN)$。

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代码实现 - 滑动窗口1

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        int ans = 0;
        int left = 0, right = 0;
        int sum = nums[0];
        while (right < nums.length) {
            if (sum < s && right < nums.length - 1) {
                sum += nums[++right];
            } else if (sum >= s) {
                ans = ans == 0 ? right - left + 1 : Math.min(ans, right - left + 1);
                sum -= nums[left++];
            } else {
                break;	// 当右端点已经在数组的末尾且sum仍比s小时，可以直接跳出
            }
        }
        
        return ans;
    }
}

代码实现 - 滑动窗口2

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int ans = Integer.MAX_VALUE;
        int sum = 0;
        int left = 0;
        
        for (int right = 0; right < nums.length; right++) {
            sum += nums[right];
            while (sum >= s) {
                ans = Math.min(ans, right - left + 1);
                sum -= nums[left++];
            }
        }
        
        return ans == Integer.MAX_VALUE ? 0 : ans;
    }
}

代码实现 - 二分查找

class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        int ans = Integer.MAX_VALUE;
        int[] sum = new int[nums.length];
        
        for (int i = 0; i < nums.length; i++) {
            sum[i] = i == 0 ? nums[i] : sum[i - 1] + nums[i];
            // 先处理掉特殊情况
            if (sum[i] >= s) {
                ans = Math.min(ans, i + 1);
            }
        }
        
        // 二分查找符合条件的点
        for (int i = 0; i < nums.length; i++) {
            int target = s + sum[i];
            int left = i + 1, right = nums.length - 1;
            while (left < right) {
                int mid = (right - left) / 2 + left;
                if (sum[mid] < target) {
                    left = mid + 1;
                } else {
                    ans = Math.min(ans, mid - i);
                    right = mid;
                }
            }
            if (sum[right] >= target) {
                ans = Math.min(ans, right - i);
            }
        }
        
        return ans == Integer.MAX_VALUE ? 0 : ans;
    }
}