[LeetCode]Minimum Size Subarray Sum

Question

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

本题难度Medium。有 种算法分别是：

题意

找出最小子串（不是子顺串subsequence），其和大于等于s

双指针法

复杂度

时间 O(N) 空间 O(1)

思路

利用左右两个指针，令

sum

为左右指针范围内的求和。

如果

sum<s

，右指针继续向右移动；

如果

sum>=s

，左指针向右移动，直到

sum<s

。用贪心法对最小长度值进行更新

附

ans

的初始值设为

nums.length+1

是very graceful（better than

-1

）

代码

public class Solution {
public int minSubArrayLen(int s, int[] nums) {
//require
int sum=0,ans=nums.length+1;
//invariant
for(int i=0,j=0;i<nums.length;i++){
sum+=nums[i];
while(sum>=s&&j<=i){
ans=Math.min(ans,i-j+1);
//move the left point
sum-=nums[j];
j++;
}
}
//ensure
return ans==(nums.length+1)?0:ans;
}
}