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  • leetcode -- Interleaving String

    Given s1s2s3, find whether s3 is formed by the interleaving of s1 and s2.

    For example,
    Given:
    s1 = "aabcc",
    s2 = "dbbca",

    When s3 = "aadbbcbcac", return true.
    When s3 = "aadbbbaccc", return false.

    [解题思路]

    DP

    Let F(i, j) denote if s1 of length i and s2 of length j could form s3 of lenght i+j, then we have four cases:
    (1) F(i, j) = F(i-1, j)                              if s1[i-1] == s3[i+j-1] && s2[j-1] != s3[i +j -1]
    (2) F(i, j) = F(i, j-1)                              if s1[i-1] != s3[i+j-1]  && s2[j-1] == s3[i +j -1]
    (3) F(i, j) = F(i-1, j) || F(i, j-1)             if s1[i-1] == s3[i+j-1] && s2[j-1] == s3[i +j -1]
    (4) F(i, j) = false                                   if s1[i-1] != s3[i+j-1] && s2[j-1] != s3[i +j -1]

     1 public class Solution {
     2     public boolean isInterleave(String s1, String s2, String s3) {
     3         // Start typing your Java solution below
     4         // DO NOT write main() function
     5         int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();
     6         if(len3 != len1 + len2){
     7             return false;
     8         }
     9         
    10         boolean[][] match = new boolean[len1 + 1][len2 + 1];
    11         match[0][0] = true;
    12         
    13         // j == 0
    14         int i = 1;
    15         while(i <= len1 && s1.charAt(i-1) == s3.charAt(i-1)){
    16             match[i][0] = true;
    17             i++;
    18         }
    19         
    20         // i == 0
    21         int j = 1;
    22         while(j <= len2 && s2.charAt(j-1) == s3.charAt(j-1)){
    23             match[0][j] = true;
    24             j++;
    25         }
    26         
    27         for(int m = 1; m <= len1; m++){
    28             for(int n = 1; n <= len2; n++){
    29                 char c = s3.charAt(m + n - 1);
    30                 if(c == s1.charAt(m - 1)){
    31                     match[m][n] |= match[m-1][n];
    32                 }
    33                 if(c == s2.charAt(n - 1)){
    34                     match[m][n] |= match[m][n-1];
    35                 }
    36             }
    37         }
    38         return match[len1][len2];
    39     }
    40 }

     ref:http://fisherlei.blogspot.com/2012/12/leetcode-interleaving-string.html

    
    
    
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  • 原文地址:https://www.cnblogs.com/feiling/p/3296057.html
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