【LeetCode】字符串系列（其他）

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38. Count and Say

题目：1、11、21、1211、111221、……

思路：从1到n依次获得

public class Solution {    public String countAndSay(int n) {        String ret = new String("1");        for(int i = 1; i < n; i++){            String pre = new String(ret);            StringBuilder sb = new StringBuilder();            int j = pre.length()-1;            while(j >= 0){                int count = 1;                while(j > 0 && pre.charAt(j) == pre.charAt(j-1)){                    count++;                    j--;                }                                sb.insert(0,pre.charAt(j--));                sb.insert(0,count);            }            ret = sb.toString();        }        return ret;    }}

58. Length of Last Word

题目：找到给定字符串最后一个单词的长度

思路：倒着做，很简单

public class Solution {    public int lengthOfLastWord(String s) {        if(s.length() == 0) return 0;        int count = 0;        int i = s.length()-1;        while(i >= 0 && s.charAt(i) == ' ') {            i--;        }        while(i >= 0 && s.charAt(i) != ' '){            count++;            i--;        }        return count;    }}

71. Simplify Path

题目：对输入路径进行简化

思路：栈——注意要讨论所有情况

public class Solution {    public String simplifyPath(String path) {        Stack<String> stack = new Stack<>();        int len = path.length();        int i = 0;        while(i < path.length()){            if(path.charAt(i) =='/'){                String sub = path.substring(i+1, len);                int j = sub.indexOf('/');                if(j == -1){                    if(!sub.equals("") && !sub.equals(".") && !sub.equals("..")) stack.push(sub);                    if(sub.equals("..") && !stack.isEmpty()) stack.pop();                     break;                }                else{                    String s = path.substring(i+1, i+1+j);                    if(!s.equals("") && !s.equals(".")){                        if(s.equals("..")){                            if(!stack.isEmpty()) stack.pop();                        }                        else stack.push(s);                    }                    i = i+1+j;                }            }        }        String ret = new String();        if(stack.isEmpty()) return "/";        while(!stack.isEmpty()){            ret = "/"+stack.pop()+ret;        }        return ret;    }}

利用split来划分会更简单

public String simplifyPath(String path) {    Deque<String> stack = new LinkedList<>();    Set<String> skip = new HashSet<>(Arrays.asList("..",".",""));    for (String dir : path.split("/")) {        if (dir.equals("..") && !stack.isEmpty()) stack.pop();        else if (!skip.contains(dir)) stack.push(dir);    }    String res = "";    for (String dir : stack) res = "/" + dir + res;    return res.isEmpty() ? "/" : res;}

72. Edit Distance

题目：找到使两个字符串相同的最小步数（通过增删改）

思路：动态规划，想清楚状态就很简单了

public class Solution {    public int minDistance(String word1, String word2) {        int n = word1.length();        int m = word2.length();        int[][] dp = new int[m+1][n+1];        dp[0][0] = 0;        for(int i = 1; i < m+1; i++){            dp[i][0] = i;        }        for(int j = 1; j < n+1; j++){            dp[0][j] = j;        }        for(int i = 1; i < m+1; i++){            for(int j = 1; j < n+1; j++){                if(word1.charAt(j-1) != word2.charAt(i-1)){                    dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1])+1;                }                else{                    dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1])+1, dp[i-1][j-1]);                }            }        }        return dp[m][n];    }}

583. Delete Operation for Two Strings

题目：只通过删除操作使两个字符串相等

思路：与72题方法完全一致，只是当字符相等时，简化了判断过程，因为只能通过删除操作。

public class Solution {    public int minDistance(String word1, String word2) {        int m = word1.length();        int n = word2.length();        int[][] dp = new int[m+1][n+1];        dp[0][0] = 0;        for(int i = 1; i <= m; i++){            dp[i][0] = i;        }        for(int j = 1; j <= n; j++){            dp[0][j] = j;        }        for(int i = 1; i <= m; i++){            for(int j = 1; j <= n; j++){                if(word1.charAt(i-1) == word2.charAt(j-1)){                    dp[i][j] = dp[i-1][j-1];                }                else{                    dp[i][j] = Math.min(Math.min(dp[i-1][j], dp[i][j-1])+1, dp[i-1][j-1]+2);                }            }        }        return dp[m][n];    }}

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