LeetCode 209 Minimum Size Subarray Sum 最小子数组和问题

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguoussubarray 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(nlog n). 

 

思路:使用滑动窗口,窗口右侧先向右扩张，使用一个变量sum来记录窗口中数字的总和，当sum大于s的时候,窗口左侧右移来缩小窗口并将sum减去nums[left]。重复此操作直到right到达数组的末尾，而left到达临界值。

 

 

public class Minimum_Size_Subarray_Sum {
    public static int minSubArrayLen(int s, int[] nums) {
        if(null==nums||nums.length==0){
            return 0;
        }
        
        int left = 0;
        int minLeft = 0;
        int minLen = nums.length;
        int sum = 0;
        int maxSum = Integer.MIN_VALUE;
        
        for(int right=0;right<nums.length;right++){
            if(sum<s){
                sum += nums[right];
                if(sum>maxSum){
                    maxSum = sum;
                }
            }
            
            while(sum>=s){
                if(right-left+1<minLen){
                    minLeft = left;
                    minLen = right - left + 1;
                }
                
                sum -= nums[left];
                left++;
            }
        }
        
        if(maxSum<s){
            return 0;
        }
        
        return minLen;
    }
    
    public static void main(String[] args) {
        int[] nums = {1,1,1,1,1,1};
        System.out.println(minSubArrayLen(6, nums));
        
    }
}

类似题目:

   窗口最小子串问题  https://www.cnblogs.com/blzm742624643/p/10357757.html