leetcode之97理解

来源：互联网发布：老婆孩子在天堂知乎编辑：程序博客网时间：2024/06/04 18:52

之前看了很多方法，理解不透，故按照个人理解加深选出几个方法，以供参考。

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        if (s3.length() != s1.length() + s2.length()) {            return false;        }        if (s1.length() == 0) return s2.equals(s3);        if (s2.length() == 0) return s1.equals(s3);                boolean sub1 = false;        boolean sub2 = false;                if (s1.charAt(0) == s3.charAt(0)) {            sub1 = isInterleave(s1.substring(1), s2, s3.substring(1));        }        if (s2.charAt(0) == s3.charAt(0)) {            sub2 = isInterleave(s1, s2.substring(1), s3.substring(1));        }                return sub1 || sub2;    }}

效率太低，报Time Limit Exceeded

public class Solution {    private boolean[][] visited;        public boolean isInterleave(String s1, String s2, String s3) {        if (visited == null) visited = new boolean[s1.length() + 1][s2.length() + 1];        if (s3.isEmpty()) return s1.isEmpty() && s2.isEmpty();                if (visited[s1.length()][s2.length()]) return false; // here                if (!s1.isEmpty() && s1.charAt(0) == s3.charAt(0)             && isInterleave(s1.substring(1), s2, s3.substring(1))) return true;        if (!s2.isEmpty() && s2.charAt(0) == s3.charAt(0)            && isInterleave(s1, s2.substring(1), s3.substring(1))) return true;                    visited[s1.length()][s2.length()] = true; //and here                return false;    }}

DFS，这样就减少了重复计算的时间， 1ms。只要在visited数组中标记过的，均为s1和s2均不匹配的情况，直接返回false

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

DP，matrix[i][j]表示当判断到s1的第i-1个字符和s2的第j个字符时，s1的第i个字符是否和s3的i+j-1个字符匹配，或者是当判断到s2的第j-1个字符和s1的第i个字符时，s2的第j个字符是否和s3的i+j-1个字符匹配。DP理解不深，今后加强练习。

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