java基础

时间 2021-01-02

标签 java 算法 code leetcode rem 字符串 get string hash 栏目 Java 繁體版

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五个问题，一次解决，字符子串问题总结

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我发如今leetcode的问题中，至少有5个子字符串寻找问题能够用滑动窗口算法解决，所以我在这里总结了这类算法的模版，但愿能够帮助你。算法

1）模版

public class Solution {
    public List<Integer> slidingWindowTemplateByHarryChaoyangHe(String s, String t) {     
        //根据问题，初始化一个储存结果的容器
        List<Integer> result = new LinkedList<>();
        if (t.length()> s.length()) return result;    
        //建立一个hashmap来保存目标子串中的字符
        //(K, V) = (Character, Frequence of the Characters)
        //key是字符， value是该字符出现的次数
        Map<Character, Integer> map = new HashMap<>();
        //将目标子串转为map存储
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c, 0) + 1);
        }        
        //维护一个计数器，去检查是否匹配目标字符串
        int counter = map.size();//必须是map的长度，不是字符串的长度是由于可能元素有重复。        
        //两个点，窗口的左端点和右端点
        int begin = 0, end = 0;       
          //匹配目标字符串的子字符串的长度
        int len = Integer.MAX_VALUE;    
        //从源字符串循环
        while (end < s.length()) {           
            char c = s.charAt(end);//获得右端点处的字符            
            if (map.containsKey(c)) {
                map.put(c, map.get(c)-1);//加一或减一
                if (map.get(c) == 0) counter--;//根据不一样的条件修改计数器
            }
            end++;            
            //increase begin pointer to make it invalid/valid again
            //计数器条件：不一样的问题选择不一样的条件
            while (counter == 0) {                
                char tempc = s.charAt(begin);//注意：选择字符是在开始端点而不是结束端点
                if (map.containsKey(tempc)) {
                    map.put(tempc, map.get(tempc) + 1);//加减一
                    if (map.get(tempc) > 0) counter++;//根据不一样的需求修改计数器
                }                
                /* save / update(min/max) the result if find a target*/
                //若是发现一个目标，保存、更新（最小、最大）结果
                // result collections or result int value                
                begin++;
            }
        }
        return result;
    }
}

3）具体问题如何应用模版

438.找到字符串中全部字母异位词
https://leetcode-cn.com/probl...leetcode

public class Solution {
    public List<Integer> findAnagrams(String s, String t) {
        List<Integer> result = new LinkedList<>();
        if(t.length()> s.length()) return result;
        Map<Character, Integer> map = new HashMap<>();
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        int counter = map.size();
        
        int begin = 0, end = 0;
        int head = 0;
        int len = Integer.MAX_VALUE;        
        
        while (end < s.length()) {
            char c = s.charAt(end);
            if (map.containsKey(c)) {
                map.put(c, map.get(c) - 1);
                if (map.get(c) == 0) counter--;
            }
            end++;
            //counter等于0意味着，end以前至少有可以凑出target的字母数量
            while (counter == 0) {
                char tempc = s.charAt(begin);
                if (map.containsKey(tempc)) {
                    map.put(tempc, map.get(tempc) + 1);
                    if(map.get(tempc) > 0){
                        counter++;
                    }
                }
                if (end - begin == t.length()) {
                    result.add(begin);
                }
                begin++;
            }            
        }
        return result;
    }
}

76. 最小覆盖子串rem

https://leetcode-cn.com/probl...字符串

public class Solution {
    public String minWindow(String s, String t) {
        if(t.length()> s.length()) return "";
        Map<Character, Integer> map = new HashMap<>();
        for(char c : t.toCharArray()){
            map.put(c, map.getOrDefault(c,0) + 1);
        }
        int counter = map.size();
        
        int begin = 0, end = 0;
        int head = 0;
        int len = Integer.MAX_VALUE;
        
        while(end < s.length()){
            char c = s.charAt(end);
            if( map.containsKey(c) ){
                map.put(c, map.get(c)-1);
                if(map.get(c) == 0) counter--;
            }
            end++;
            
            while(counter == 0){
                char tempc = s.charAt(begin);
                if(map.containsKey(tempc)){
                    map.put(tempc, map.get(tempc) + 1);
                    if(map.get(tempc) > 0){
                        counter++;
                    }
                }
                if(end-begin < len){
                    len = end - begin;
                    head = begin;
                }
                begin++;
            }
            
        }
        if(len == Integer.MAX_VALUE) return "";
        return s.substring(head, head+len);
    }
}

3.无重复字符的最长子串get

https://leetcode-cn.com/probl...string

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        Map<Character, Integer> map = new HashMap<>();
        int begin = 0, end = 0, counter = 0, d = 0;

        while (end < s.length()) {
            // > 0 means repeating character
            //if(map[s.charAt(end++)]-- > 0) counter++;
            char c = s.charAt(end);
            map.put(c, map.getOrDefault(c, 0) + 1);
            if(map.get(c) > 1) counter++;
            end++;
            
            while (counter > 0) {
                //if (map[s.charAt(begin++)]-- > 1) counter--;
                char charTemp = s.charAt(begin);
                if (map.get(charTemp) > 1) counter--;
                map.put(charTemp, map.get(charTemp)-1);
                begin++;
            }
            d = Math.max(d, end - begin);
        }
        return d;
    }
}

30.串联全部单词的子串hash

https://leetcode-cn.com/probl...it

public class Solution {
    public List<Integer> findSubstring(String S, String[] L) {
        List<Integer> res = new LinkedList<>();
        if (L.length == 0 || S.length() < L.length * L[0].length())   return res;
        int N = S.length();
        int M = L.length; // *** length
        int wl = L[0].length();
        Map<String, Integer> map = new HashMap<>(), curMap = new HashMap<>();
        for (String s : L) {
            if (map.containsKey(s))   map.put(s, map.get(s) + 1);
            else                      map.put(s, 1);
        }
        String str = null, tmp = null;
        for (int i = 0; i < wl; i++) {
            int count = 0;  // remark: reset count 
            int start = i;
            for (int r = i; r + wl <= N; r += wl) {
                str = S.substring(r, r + wl);
                if (map.containsKey(str)) {
                    if (curMap.containsKey(str))   curMap.put(str, curMap.get(str) + 1);
                    else                           curMap.put(str, 1);
                    
                    if (curMap.get(str) <= map.get(str))    count++;
                    while (curMap.get(str) > map.get(str)) {
                        tmp = S.substring(start, start + wl);
                        curMap.put(tmp, curMap.get(tmp) - 1);
                        start += wl;
                        
                        //the same as https://leetcode.com/problems/longest-substring-without-repeating-characters/
                        if (curMap.get(tmp) < map.get(tmp)) count--;
                        
                    }
                    if (count == M) {
                        res.add(start);
                        tmp = S.substring(start, start + wl);
                        curMap.put(tmp, curMap.get(tmp) - 1);
                        start += wl;
                        count--;
                    }
                }else {
                    curMap.clear();
                    count = 0;
                    start = r + wl;//not contain, so move the start
                }
            }
            curMap.clear();
        }
        return res;
    }
}

java基础

五个问题，一次解决，字符子串问题总结

1）模版

2）相关问题

3）具体问题如何应用模版