LeetCode 题解（126）: Minimum Size Subarray Sum

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.

题解：

双指针。

C++版：

class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        if(nums.size() == 0)
            return 0;
        
        int i = 0, j = 0;
        int sum = 0;
        while(sum < s && j < nums.size()) {
            sum += nums[j++];
        }
        
        if(j == nums.size()) {
            if(sum < s)
                return 0;
            else if(sum == s)
                return nums.size();
        }

        int global = j - i;
        j--;
        while(i != j) {
            if(nums[j] > nums[i] && sum > s) {
                sum -= nums[i++];
                if(sum >= s)
                    if(j - i + 1 < global)
                        global = j - i + 1;
            } else {
                if(j + 1 < nums.size()) {
                    sum -= nums[i++];
                    sum += nums[++j];
                    if(sum >= s)
                        if(j - i + 1 < global)
                            global = j - i + 1;
                } else {
                    break;
                }
            }
        }
        return global;
    }
};

Java版：

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums.length == 0)
            return 0;
            
        int i = 0, j = 0, sum = 0, result = nums.length + 1;
        while(j < nums.length) {
            while(j < nums.length && sum < s) {
                sum += nums[j++];
            }
            while(sum >= s && i < j) {
                result = Math.min(result, j - i);
                sum -= nums[i++];
            }
        }
        return result == nums.length + 1? 0 : result;
    }
}

Python版：

class Solution:
    # @param {integer} s
    # @param {integer[]} nums
    # @return {integer}
    def minSubArrayLen(self, s, nums):
        i, j, sum, result = 0, 0, 0, len(nums) + 1
        if len(nums) == 0:
            return 0
            
        while j < len(nums):
            while j < len(nums) and sum < s:
                sum += nums[j]
                j += 1
            while i < j and sum >= s:
                result = min(result, j - i)
                sum -= nums[i]
                i += 1
                
        return result if result != len(nums) + 1 else 0