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.
一刷
题解：
首先从起点j开始加，一旦超过sum, 则从头部开始减去已经加上的。直至小于sum, 在这个过程中不停地update min_len
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
if(nums == null || nums.length == 0) return 0;
int i=0, j=0, sum = 0, min = Integer.MAX_VALUE;
while(j<nums.length){
sum += nums[j];
j++;
while(sum>=s){
min = Math.min(min, j-i);
sum -= nums[i];
i++;
}
}
return min == Integer.MAX_VALUE? 0 : min;
}
}
二刷
思路同上
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
if(nums == null || nums.length == 0) return 0;
int i=0, j=0, sum = 0, min = Integer.MAX_VALUE;
while(j<nums.length){
sum += nums[j];
j++;
while(sum>=s){
min = Math.min(j-i, min);//not include j
sum -= nums[i];
i++;
}
}
return min==Integer.MAX_VALUE? 0:min;
}
}
