Java算法练习题,每天进步一点点(1)

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        if (l1 > l2 || "".equals(s1) || "".equals(s2)) {
            return false;
        }
        //创建个need数组，表示所需要的字符以及个数，通过遍历s1的得到
        int[] need = new int[26];
        for (int i = 0; i < l1; i++) {
            need[s1.charAt(i) - 'a']++;
        }
        //滑动窗口
        int l = 0, r = 0;
        //如果l=l2-l1就可以停了，后面的长度都不够了
        while (l <= l2 - l1) {
            //如果符合条件，即need[s2.charAt(r) - 'a'] > 0，就是当前窗口右端碰到的的是需要的字符
            while (r < l + l1 && need[s2.charAt(r) - 'a'] > 0) {
                //更新所需的字符个数
                need[s2.charAt(r) - 'a']--;
                //扩大窗口范围
                r++;
            }
            //找到所符合的个数了，就是需要的子串已经找到了
            if (r == l + l1) {
                return true;
            }
            //移动左窗口，这样左边的字符从窗口中退出来了，就需要把need[s2.charAt(l) - 'a']++，维护need
            need[s2.charAt(l) - 'a']++;
            //移动左边的指针
            l++;
        }
        return false;
    }
}

package com.keafmd.Likou.Day0729;
import java.util.Arrays;
import java.util.HashMap;
/**
 * Keafmd
 *
 * @ClassName: StringArrangement
 * @Description: https://leetcode-cn.com/problems/permutation-in-string/
 * @author: 牛哄哄的柯南
 * @date: 2021-07-29 9:11
 */
public class StringArrangement {
    public static void main(String[] args) {
        String s1 = "hello", s2 = "ooolleooolleh";
        boolean b = new StringArrangementSolution().checkInclusion(s1, s2);
        System.out.println(b);
    }
}
class StringArrangementSolution {
    public boolean checkInclusion(String s1, String s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        if (l1 > l2 || "".equals(s1) || "".equals(s2)) {
            return false;
        }
        //创建个need数组，表示所需要的字符以及个数，通过遍历s1的得到
        int[] need = new int[26];
        for (int i = 0; i < l1; i++) {
            need[s1.charAt(i) - 'a']++;
        }
        //滑动窗口
        int l = 0, r = 0;
        //如果l=l2-l1就可以停了，后面的长度都不够了
        while (l <= l2 - l1) {
            //如果符合条件，即need[s2.charAt(r) - 'a'] > 0，就是当前窗口右端碰到的的是需要的字符
            while (r < l + l1 && need[s2.charAt(r) - 'a'] > 0) {
                //更新所需的字符个数
                need[s2.charAt(r) - 'a']--;
                //扩大窗口范围
                r++;
            }
            //找到所符合的个数了，就是需要的子串已经找到了
            if (r == l + l1) {
                return true;
            }
            //移动左窗口，这样左边的字符从窗口中退出来了，就需要把need[s2.charAt(l) - 'a']++，维护need
            need[s2.charAt(l) - 'a']++;
            //移动左边的指针
            l++;
        }
        return false;
    }
}
class StringArrangementSolution2 {
    public boolean checkInclusion(String s1, String s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        if (s1 == null || s2 == null || l1 > l2 || s1 == "" || s2 == "") {
            return false;
        }
        int[] need = new int[26];
        for (int i = 0; i < l1; i++) {
            need[s1.charAt(i) - 'a']--;
        }
        int l = 0, r = 0;
        int count = 0;
        while (r < l2) {
            int x = s2.charAt(r) - 'a';
            need[x]++;
            while (need[x] > 0) {
                need[s2.charAt(l) - 'a']--;
                l++;
            }
            if (r - l + 1 == l1) {
                return true;
            }
            r++;
        }
        return false;
    }
}

class StringArrangementSolution1 {
    public boolean checkInclusion(String s1, String s2) {
        int l1 = s1.length();
        int l2 = s2.length();
        if (s1 == null || s2 == null || l1 > l2 || s1 == "" || s2 == "") {
            return false;
        }
        int[] num1 = new int[26];
        int[] num2 = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            num1[s1.charAt(i) - 'a']++;
            num2[s2.charAt(i) - 'a']++;
        }
        if (Arrays.equals(num1, num2)) {
            return true;
        }

        int l = 0, r = 0;
        int count = 0;
        for (int i = l1; i < l2; i++) {
            num2[s2.charAt(i) - 'a']++;
            num2[s2.charAt(i - l1) - 'a']--;
            if (Arrays.equals(num1, num2)) {
                return true;
            }
        }
        return false;
    }
}