LeetCode--97. Interleaving String

最新推荐文章于 2024-08-30 08:34:12 发布

原创最新推荐文章于 2024-08-30 08:34:12 发布 · 234 阅读

0 ·

CC 4.0 BY-SA版权

leetcode与剑指Offer 专栏收录该内容

222 篇文章

订阅专栏

本文深入探讨了交错字符串问题，提供了五种不同的算法解决方案，包括暴力递归、记忆化搜索、二维DP、一维DP和BFS算法。每种算法都附有详细的实现代码，帮助读者理解如何判断一个字符串是否可以通过交错两个输入字符串得到。

97. Interleaving String
这是一道极具代表性的字符串动态规划算法题
算法一：暴力递归

public static boolean isInterleave(String s1, String s2, String s3) {
        return recursive(s1, 0, s2, 0, "", s3);
    }

    public static boolean recursive(String s1, int i, String s2, int j, String res, String s3) {

        if (res.equals(s3) && (i == s1.length()) && (j == s2.length())) {
            return true;
        }
        boolean ans = false;
        if (i < s1.length())
            ans |= recursive(s1, i + 1, s2, j, res + s1.charAt(i), s3);
        if (j < s2.length())
            ans |= recursive(s1, i, s2, j + 1, res + s2.charAt(j), s3);
        return ans;
    }

算法二：记忆化搜索

public static boolean isInterleave1(String s1, String s2, String s3) {

        int m = s1.length(), n = s2.length();
        int[][] memo = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++)
                memo[i][j] = -1;
        }
        return recursive1(s1,0,s2,0,s3,0,memo);
    }

    public static boolean recursive1(String s1, int i, String s2, int j, String s3, int k, int[][] memo) {

        if (i == s1.length())
            return s3.substring(k).equals(s2.substring(j));
        if (j == s2.length())
            return s3.substring(k).equals(s1.substring(i));
        if (memo[i][j] >= 0)
            return memo[i][j] == 1 ? true : false;
        boolean ans = false;
        if ((s1.charAt(i) == s3.charAt(k) && recursive1(s1, i + 1, s2, j, s3, k + 1, memo)) || (s2.charAt(j) == s3.charAt(k) && recursive1(s1, i, s2, j + 1, s3, k+1, memo))) {
            ans = true;
        }
        memo[i][j] = ans ? 1 : 0;
        return ans;
    }

算法三：二维DP

public static boolean isInterleave2(String s1, String s2, String s3) {

        //dp[i][j]
        int m=s1.length(),n=s2.length();
        boolean[][] dp=new boolean[m+1][n+1];
        for(int i=0;i<=m;i++){
            for(int j=0;j<=n;j++)
            {
                if(i==0 && j==0)
                    dp[i][j]=true;
                else if(i==0){
                    dp[i][j]=dp[i][j-1] && s2.charAt(j-1)==s3.charAt(i+j-1);
                }
                else if(j==0){
                    dp[i][j]=dp[i-1][j] && s1.charAt(i-1)==s3.charAt(i+j-1);
                }
                else
                    dp[i][j]=(dp[i][j-1] && s2.charAt(j-1)==s3.charAt(i+j-1)) ||(dp[i-1][j] && s1.charAt(i-1)==s3.charAt(i+j-1));
            }
        }
        return dp[m][n];
    }

算法四：

public class Solution {
    public boolean isInterleave(String s1, String s2, String s3) {
        if (s3.length() != s1.length() + s2.length()) {
            return false;
        }
        boolean dp[] = new boolean[s2.length() + 1];
        for (int i = 0; i <= s1.length(); i++) {
            for (int j = 0; j <= s2.length(); j++) {
                if (i == 0 && j == 0) {
                    dp[j] = true;
                } else if (i == 0) {
                    dp[j] = dp[j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
                } else if (j == 0) {
                    dp[j] = dp[j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
                } else {
                    dp[j] = (dp[j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) || (dp[j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
                }
            }
        }
        return dp[s2.length()];
    }
}

算法五：BFS

static class Coordinate{
        int x,y;

        public Coordinate(int x,int y){
            this.x=x;
            this.y=y;
        }
    }

    public static boolean isInterleave3(String s1, String s2, String s3){

        int m=s1.length(),n=s2.length();
        boolean[][] visited=new boolean[m+1][n+1];
        Queue<Coordinate> q=new LinkedList<>();
        q.add(new Coordinate(0,0));
        while (!q.isEmpty()){

            Coordinate tmp=q.poll();
            int x=tmp.x,y=tmp.y;
            if(x==m && y==n)
                return true;
            int down=x+1;
            if(down<=s1.length() && s1.charAt(down-1)==s3.charAt(down+y-1) && !visited[down][y])
            {
                visited[down][y]=true;
                q.add(new Coordinate(down,y));
            }

            int right=y+1;
            if(right<=s2.length() && s2.charAt(right-1)==s3.charAt(x+right-1) && !visited[x][right]){
                visited[x][right]=true;
                q.add(new Coordinate(x,right));
            }
        }
        return false;
    }