LeetCode-Interleaving String

作者：disappearedgod

文章出处：http://blog.csdn.net/disappearedgod/article/details/23621903

时间：2014-6-13

题目

Interleaving String

 Total Accepted: 9375 Total Submissions: 50195My Submissions

Given s1, s2, s3, 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.

解法

一个有问题的想法

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        int i = 0 , j = 0 , k = 0;        char c1 , c2 , c3 ;        int len1 = s1.length(), len2 = s2.length(), len3 = s3.length();        if(len3 == 0)            return len1 == 0 && len2 == 0;        while((i<len1 || j<len2) && k < len3){            c3 = s3.charAt(k);            if(len1>i){                c1 = s1.charAt(i);                if(c3 == c1){                    i++;                    k++;                    continue;                }                else                    break;                            }            if(len2>j){                c2 = s2.charAt(j);                if (c3 == c2){                    j++;                    k++;                    continue;                }                else                    break;            }            break;                 }                return i ==len1 && j==len2  && k == len3;            }}

结果

Input:"aa", "ab", "aaba"Output:falseExpected:true

递归想法

代码

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {         if(s3==null)            return s1==null && s2==null;        int len1 = 0, len2 = 0, len3 = 0;        if(s1!=null)            len1 = s1.length();        if(s2!=null)            len2 = s2.length();        len3 = s3.length();        if(len3!=len1+len2)            return false;                if( len1 == 0)            return s2.equals(s3);        if( len2 == 0)            return s1.equals(s3);        len1--;len2--;len3--;        if(s1.charAt(0)==s3.charAt(0) && s2.charAt(0)==s3.charAt(0)){            if(len1 > 0 && len2 > 0 && len3 > 0)                return isInterleave(s1.substring(1,len1),s2,s3.substring(1,len3)) || isInterleave(s1,s2.substring(1,len2),s3.substring(1,len3));            else if(len1 > 0 && len3 > 0)// len2++ == 1                return isInterleave(s1.substring(1,len1),null,s3.substring(1,len3)) || isInterleave(s1.substring(1,len1),s2,s3.substring(1,len3));            else if(len2 > 0 && len3 > 0)//len1++ == 1                return isInterleave(null,s2,s3.substring(1,len3)) || isInterleave(s1,s2.substring(1,len2),s3.substring(1,len3));            else                return false;                                }        else if(s1.charAt(0)==s3.charAt(0)){            if(len1 > 0 && len3 > 0)                return isInterleave(s1.substring(1,len1),s2,s3.substring(1,len3));            else if(len3 > 0)                return isInterleave(null,s2,s3.substring(1,len3));            else                return len2 == 0;        }        else if(s2.charAt(0)==s3.charAt(0)){            if(len2 > 0  && len3 > 0)                return isInterleave(s1,s2.substring(1,len2),s3.substring(1,len3));            else if (len3 > 0)                return isInterleave(s1,null,s3.substring(1,len3));            else                return len1 == 0;        }        else{            return false;        }    }    }

结果

Last executed input:"bbbbbabbbbabaababaaaabbababbaaabbabbaaabaaaaababbbababbbbbabbbbababbabaabababbbaabababababbbaaababaa", "babaaaabbababbbabbbbaabaabbaabbbbaabaaabaababaaaabaaabbaaabaaaabaabaabbbbbbbbbbbabaaabbababbabbabaab", "babbbabbbaaabbababbbbababaabbabaabaaabbbbabbbaaabbbaaaaabbbbaabbaaabababbaaaaaabababbababaababbababbbababbbbaaaabaabbabbaaaaabbabbaaaabbbaabaaabaababaababbaaabbbbbabbbbaabbabaabbbbabaaabbababbabbabbab"

非递归思路

栈储存的思路

每次遇到s1,s2被指针指向的位置时候，储存相同的点。为了回退。（其实是递归的一种非递归实现，还是超时）

public class Solution {    class Node{        int x;        int y;        boolean isPop = false;        Node(int i, int j, boolean bool){            x = i;            y = j;            isPop = bool;        }    }    public boolean isInterleave(String s1, String s2, String s3) {        if(s3==null)            return s1==null && s2==null;        int len1 = 0, len2 = 0, len3 = 0;        if(s1 != null)            len1 = s1.length();        if(s2 != null)            len2 = s2.length();        len3 = s3.length();        if(len3 != len1+len2)            return false;        if( len1 == 0)            return s2.equals(s3);        if( len2 == 0)            return s1.equals(s3);        Queue<Node> q = new LinkedList<Node>();        q.offer(new Node(0, 0, false));        int i = 0, j = 0;        while(!q.isEmpty()){            Node node = q.poll();            i = node.x;            j = node.y;            boolean isPop = node.isPop;            char c1, c2, c3;            while(i + j < len3){                if(i < len1)                    c1 = s1.charAt(i);                else                    c1 = '.';                if(j < len2)                    c2 = s2.charAt(j);                else                    c2 = '.';                c3 = s3.charAt(i+j);                 if(c1 == c3 && c2 == c3){                    if(!isPop){                        q.offer(new Node(i, j, true));                        i++;                    }                    else{                        j++;                        isPop = false;                    }                }                else if(c1 == c3){                    i++;                    if( i > len1)                        return false;                }                else if(c2 == c3){                    j++;                    if(j > len2)                        return false;                }                else{                    break;                }            }        }        return i + j == len3;    }}

结果

Submission Result: Time Limit Exceeded

Last executed input:"cbcccbabbccbbcccbbbcabbbabcababbbbbbaccaccbabbaacbaabbbc", "abcbbcaababccacbaaaccbabaabbaaabcbababbcccbbabbbcbbb", "abcbcccbacbbbbccbcbcacacbbbbacabbbabbcacbcaabcbaaacbcbbbabbbaacacbbaaaabccbcbaabbbaaabbcccbcbabababbbcbbbcbb"

DP解法

想法

                     0 c a b       ccabcac
                0   1 0 0 0
                c   1 1 1 1 
                c   0 0 0 1
                a   0 0 0 1
                c   0 0 0 1
                
                     0 c a b       ccabcac
                0   1 ? 0 0
                c   ? 0 0 0 
                c   0 0 0 0
                a   0 0 0 0
                c   0 0 0 0
                
                     0    c   a   b       ccabcac
                0   1    0   0   0
                c   1    1   1   1 
                c   1-* 0   0   1
                a   1-* 0   0   1
                c   0    0   0   1
                

      0    c   a   b       ccabcac
                0   1    1   0   0
                c   1    1   1   1 
                c   1 0   0   1
                a   1 0   0   1
                c   0    0   0   1
                           

代码

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        int l1 = s1.length();        int l2 = s2.length();        int l3 = s3.length();        if(l1 != l3 -l2)            return false;        boolean [][] dp = new boolean[l1 + 1][l2 + 1];        dp[0][0] = false;        for(int k = 0; k < l3; k++){            for(int i = 0; i < l3 && i < l1; i++){                int j = k - i;                if(j >= l2)                    continue;                dp[i + 1][j + 1] = false;                if(s1.charAt(i) == s3.charAt(k) && dp[i][j + 1])                    dp[i + 1][j + 1] = true;                if(s2.charAt(j) == s3.charAt(k) && dp[i + 1][j])                    dp[i + 1][j + 1] = true;            }                    }        return dp[l1][l2];    }}

结果

数组越界

改正

递推关系应该是由前面推向后面，所以要做到正好。

代码 

public class Solution {    public boolean isInterleave(String s1, String s2, String s3) {        int l1 = s1.length();        int l2 = s2.length();        int l3 = s3.length();        if(l1 != l3 -l2)            return false;        boolean [][] dp = new boolean[l1 + 1][l2 + 1];        dp[0][0] = true;        for(int k = 1; k <= l3; k++){            for(int i = 0; i <= k && i <= l1; i++){                int j = k - i;                if(j > l2)                    continue;                dp[i][j] = false;                if(i>0 && s1.charAt(i-1) == s3.charAt(k-1) && dp[i-1][j])                    dp[i][j] = true;                if(j > 0 && s2.charAt(j-1) == s3.charAt(k-1) && dp[i][j-1])                    dp[i][j] = true;            }                    }        return dp[l1][l2];    }}

结果

Submit TimeStatusRun TimeLanguage3 minutes agoAccepted444 msjava

参考

[LeetCode] Interleaving String

结果

返回

LeetCode Solution(持续更新，java>c++)