[leetcode]Minimum Size Subarray Sum

来源：互联网发布：unity3d树木高模编辑：程序博客网时间：2024/05/19 18:16

题目描述如下：

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.

题目大意就是寻找到最短的连续子串和大于等于给定的数。

此类题目之前有接触过，一般用的是滑动窗口的方法，即两个指针相互行进，扫描一遍数组后得出答案。

第一版代码如下:

public class Solution {    public int minSubArrayLen(int s, int[] nums) {        int res = Integer.MAX_VALUE;        int arrLen = nums.length;        int first = 0, second = 0;        int currSum = 0;        boolean flag = true;        while(first < arrLen){            if(flag){                currSum += nums[first];                flag = false;            }            if(currSum >= s){                if(res > first - second + 1){                    res = first - second + 1;                }                currSum -= nums[second];                second ++;            }else{                first ++;                flag = true;            }        }            return res;    }}

这样的做法超时，也就是说每次循环若只移动两个指针中的其一一格是不够的。于是进行优化，即在一次循环中，让前后指针在给定的范围内不断移动，这样就可以减少判断的次数。修改代码如下：

public class Solution {    public int minSubArrayLen(int s, int[] nums) {        if(nums.length == 0) return 0;        int res = Integer.MAX_VALUE;        int arrLen = nums.length;        int first = 0, second = 0;        int currSum = 0;        while(first < arrLen && second < arrLen){                while(currSum < s && first < arrLen){                    currSum += nums[first];                    first++;                }                while(currSum >= s && second < first){                    if(res > first - second){                        res = first - second;                    }                    currSum -= nums[second];                    second ++;                }            }            return res;    }}

但是光做到这样还是不行，还需要对特殊条件进行判断：

1、输入的数组为空时，返回的值是0；

2、当数组无法满足给定数值时，返回的也是0；

加上对特殊条件的鉴别，最终代码如下：

public class Solution {    public int minSubArrayLen(int s, int[] nums) {        if(nums.length == 0) return 0;        int res = Integer.MAX_VALUE;        int arrLen = nums.length;        int first = 0, second = 0;        int currSum = 0;        while(first < arrLen && second < arrLen){                while(currSum < s && first < arrLen){                    currSum += nums[first];                    first++;                }                while(currSum >= s && second < first){                    if(res > first - second){                        res = first - second;                    }                    currSum -= nums[second];                    second ++;                }            }            return res == Integer.MAX_VALUE ? 0 : res;    }}

题目链接：https://leetcode.com/problems/minimum-size-subarray-sum/

0 0