209. Minimum Size Subarray Sum**

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.

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

总结：O(N^2)

public class Solution {
public int minSubArrayLen(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 low, int high, int key, int[] sums){
while(low<=high){
int mid = low +(high-low)/2;
if(sums[mid] >=key){
high = mid -1;
}
else{
low = mid +1;
}
}
return low;
}
}

总结：O(nlogn)