mirror of
https://github.com/chefyuan/algorithm-base.git
synced 2024-11-28 14:58:55 +00:00
152 lines
4.7 KiB
Java
152 lines
4.7 KiB
Java
> 如果阅读时，发现错误，或者动画不可以显示的问题可以添加我微信好友 **[tan45du_one](https://raw.githubusercontent.com/tan45du/tan45du.github.io/master/个人微信.15egrcgqd94w.jpg)** ，备注 github + 题目 + 问题 向我反馈
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>
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> 感谢支持，该仓库会一直维护，希望对各位有一丢丢帮助。
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>
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> 另外希望手机阅读的同学可以来我的 <u>[**公众号：袁厨的算法小屋**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u> 两个平台同步，想要和题友一起刷题，互相监督的同学，可以在我的小屋点击<u>[**刷题小队**](https://raw.githubusercontent.com/tan45du/test/master/微信图片_20210320152235.2pthdebvh1c0.png)</u>进入。
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#### [209. 长度最小的子数组](https://leetcode-cn.com/problems/minimum-size-subarray-sum/)
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我们下面再看一种新类型的双指针，也就是我们大家熟知的滑动窗口。这也是我们做题时经常用到的，下面我们来看一下题目吧！
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#### 题目描述
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> 给定一个含有 n 个正整数的数组和一个正整数 s ，找出该数组中满足其和 ≥ s 的长度最小的 连续 子数组，并返回其长度。如果不存在符合条件的子数组，返回 0。
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示例：
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> 输入：s = 7, nums = [2,3,1,2,4,3]
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> 输出：2
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> 解释：子数组 [4,3] 是该条件下的长度最小的子数组。
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#### 题目解析
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滑动窗口：**就是通过不断调节子数组的起始位置和终止位置，进而得到我们想要的结果**，滑动窗口也是双指针的一种。
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下面我们来看一下这道题目的做题思路，其实原理也很简单，我们创建两个指针，一个指针负责在前面探路，并不断累加遍历过的元素的值，当和大于等于我们的目标值时，后指针开始进行移动，判断去除当前值时，是否仍能满足我们的要求，直到不满足时后指针 停止，前面指针继续移动，直到遍历结束。是不是很简单呀。前指针和后指针之间的元素个数就是我们的滑动窗口的窗口大小。见下图
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好啦，该题的解题思路我们已经了解啦，下面我们看一下，代码的运行过程吧。
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#### 题目代码
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Java Code:
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```java
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class Solution {
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public int minSubArrayLen(int s, int[] nums) {
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int len = nums.length;
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int windowlen = Integer.MAX_VALUE;
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int i = 0;
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int sum = 0;
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for (int j = 0; j < len; ++j) {
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sum += nums[j];
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while (sum >= s) {
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windowlen = Math.min (windowlen, j - i + 1);
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sum -= nums[i];
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i++;
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}
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}
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return windowlen == Integer.MAX_VALUE ? 0 : windowlen;
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}
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}
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```
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C++ Code:
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```cpp
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class Solution {
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public:
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int minSubArrayLen(int t, vector<int>& nums) {
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int n = nums.size();
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int i = 0, sum = 0, winlen = INT_MAX;
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for(int j = 0; j < n; ++j){
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sum += nums[j];
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while(sum >= t){
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winlen = min(winlen, j - i + 1);
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sum -= nums[i++];
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}
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}
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return winlen == INT_MAX? 0: winlen;
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}
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};
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```
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Python3 Code:
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```python
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from typing import List
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import sys
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class Solution:
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def minSubArrayLen(self, s: int, nums: List[int])->int:
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leng = len(nums)
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windowlen = sys.maxsize
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i = 0
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sum = 0
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for j in range(0, leng):
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sum += nums[j]
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while sum >= s:
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windowlen = min(windowlen, j - i + 1)
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sum -= nums[i]
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i += 1
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if windowlen == sys.maxsize:
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return 0
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else:
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return windowlen
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```
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Swift Code
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```swift
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class Solution {
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func minSubArrayLen(_ target: Int, _ nums: [Int]) -> Int {
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var sum = 0, windowlen = Int.max, i = 0
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for j in 0..<nums.count {
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sum += nums[j]
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while sum >= target {
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windowlen = min(windowlen, j - i + 1)
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sum -= nums[i]
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i += 1
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}
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}
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return windowlen == Int.max ? 0 : windowlen
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}
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}
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```
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Go Code:
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```go
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func minSubArrayLen(target int, nums []int) int {
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length := len(nums)
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winLen := length + 1
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i := 0
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sum := 0
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for j := 0; j < length; j++ {
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sum += nums[j]
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for sum >= target {
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winLen = min(winLen, j - i + 1)
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sum -= nums[i]
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i++
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}
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}
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if winLen == length + 1 {
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return 0
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}
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return winLen
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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