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.
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
if(nums==null || nums.length==0) return 0;
int len = nums.length;
int[] sum = new int[len+1];
for(int i=0; i<len; i++) sum[i+1] = sum[i] + nums[i];
if(sum[len]<s) return 0;
int ans = Integer.MAX_VALUE;
for(int i=0; i<len; i++){
for(int j=i+1; j<=len; j++){
if(sum[j]-sum[i]>=s){
if(j-i<ans){
ans = j-i;
break;
}
}
}
}
return ans;
}
}
别人的做法,用了两个指针的做法
public int minSubArrayLen(int s, int[] a) {
if (a == null || a.length == 0)
return 0;
int i = 0, j = 0, sum = 0, min = Integer.MAX_VALUE;
while (j < a.length) {
sum += a[j++];
while (sum >= s) {
min = Math.min(min, j - i);
sum -= a[i++];
}
}
return min == Integer.MAX_VALUE ? 0 : min;
}