滑动窗口Sliding Window Algorithm

应用及优点：

1.可用于解决数组或者字符串的子元素问题。

2.用单循环代替了嵌套循环问题，时间复杂度低。

3.用双指针维护动态窗口。

 

相关算法题：

Longest Substring Without Repeating Characters无重复最长子串

Find All Anagrams in a String（找到字符串中所有字母异位词）

Minimum Window Substring最小覆盖子串

209. Minimum Size Subarray Sum长度最小子数组

以“Find All Anagrams in a String”为例：

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        int [] need=new int[128];
        int [] win=new int[128];
        int left=0;int right=0;
        List<Integer> res=new ArrayList<>();
        for(int i=0;i<p.length();i++){
            need[p.charAt(i)]++;//将目标数组中各元素计数。
        }
        while(right<s.length()){
            win[s.charAt(right)]++;//窗口右移
            while(right<s.length()&&win[s.charAt(right)]>need[s.charAt(right)]){//窗口停止右移条件
                win[s.charAt(left)]--;
                left+；//窗口左移
            }
            if(right-left+1==p.length()){
                res.add(left);
            }
            right++;

        }
        return res;
    }
}

219. Contains Duplicate II

Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

Example 1:

Input: nums = [1,2,3,1], k = 3
Output: true

维护固定长度的滑动窗口+查找表

class Solution {
    public boolean containsNearbyDuplicate(int[] nums, int k) {
        Set<Integer> s=new HashSet<>();
        for(int i=0;i<nums.length;i++){
            if(s.contains(nums[i])){
                return true;
            }
            s.add(nums[i]);
            if(s.size()>k){
                s.remove(nums[i-k]);
            }
        }
        return false;
    }
}