209. Minimum Size Subarray Sum

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.

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).

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        if(nums==null||nums.length==0) return 0;
        int left = 0;
        int right = 0;
        int sum = 0;
        int count = Integer.MAX_VALUE;
        while(right<nums.length){
            sum+=nums[right];
            while(sum>=s){
                count = Math.min(count,right-left+1);
                sum-=nums[left];
                left++;
            }
            right++;
        }
        return count==Integer.MAX_VALUE?0:count;
    }
}
//the time couplexity could be O(n),the space complexity could be O(1);

public class Solution {
    public int minSubArrayLen(int s, int[] nums) {
        int[] sums = new int[nums.length+1];
        for(int i=0;i<nums.length;i++){
            sums[i+1] = sums[i]+nums[i];
        }
        int min = Integer.MAX_VALUE;
        for(int i=0;i<sums.length;i++){
            int end = binarysearch(sums,i+1,sums.length-1,sums[i]+s);
            if(end==sums.length) break;
            if(end-i<min) min = end-i;
        }
        return min==Integer.MAX_VALUE?0:min;
    }
    public int binarysearch(int[] sums,int left,int right,int target){
        while(left<=right){
            int mid = left+(right-left)/2;
            if(sums[mid]<target){
                left = mid+1;
            }else{
                right = mid-1;
            }
        }
        return left;
    }
}
// the time complexity could be nlogn,the space complexity could be O(1);