Leetcode 161. One Edit Distance

161. One Edit Distance

Total Accepted: 18099 Total Submissions: 61201 Difficulty: Medium

Given two strings S and T, determine if they are both one edit distance apart.

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 (H) Edit Distance

把人搞疯的一道题。。

思路：

本来直接按之前那个hard题目做，然后最后判断==1就行了。结果过不了大数据：

public class Solution { // 超时    public boolean isOneEditDistance(String s, String t) {        if(s.length() == 0 && t.length() == 0) return false;        if(s.length() == 0) return t.length() == 1;        if(t.length() == 0) return s.length() == 1;                char[] ss = s.toCharArray();        char[] tt = t.toCharArray();        int[][] dp = new int[ss.length + 1][tt.length + 1]; // dp[i][j] : s.substring(0, i) <-> t.substring(0, j)        // initialize        for(int i = 0; i < ss.length + 1; i++)  dp[i][0] = i;                for(int i = 0; i < tt.length + 1; i++)  dp[0][i] = i;                for(int i = 1; i < ss.length + 1; i++){            for(int j = 1; j < tt.length + 1; j++){                if(ss[i - 1] == tt[j -1])  dp[i][j] = dp[i - 1][j - 1];                else    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;            }        }        return dp[ss.length][tt.length] == 1;    }}

于是如何提前判断就是问题：

if(i == j){if(dp[i][j] > 1) return false;}

118 / 130 test cases passed.
Status: Wrong Answer
Input:
"ab"
"cab"
Output:
false
Expected:
true

if(Math.abs(i - j) <= 1){if(dp[i][j] > 1) return false;}

111 / 130 test cases passed.
Status: Wrong Answer
Input:
"cb"
"ab"
Output:
false
Expected:
true

意思是不让DP咯 

解法出自这里：

public class Solution {    public boolean isOneEditDistance(String s, String t) {        if (s == null || t == null || s.equals(t) || Math.abs(s.length() - t.length()) > 1) return false;        if (s.length() > t.length()) return isOneEditDistance(t, s);        boolean hasDiff = false;        for (int i = 0, j = 0; i < s.length(); i++, j++) {            if (s.charAt(i) != t.charAt(j)) {                if (hasDiff) return false;                hasDiff = true;                if (s.length() < t.length()) i--;            }        }        return true;    }}// 假如字符不同，则hasEdited为true，假如s比t短，则下标i退回1来继续比较insert / delete的case。否则比较的是replace。

另还有一讨论区解法，时间都是2ms： 

public class Solution {     public boolean isOneEditDistance(String s, String t) {    int m = s.length();    int n = t.length();    if (Math.abs(m - n) > 1) {    return false;    }        int len = Math.min(m, n);    for (int i = 0; i < len; i++) {    if (s.charAt(i) != t.charAt(i)) {        boolean result = s.substring(i).equals(t.substring(i + 1));    result = result || s.substring(i + 1).equals(t.substring(i));    result = result || s.substring(i + 1).equals(t.substring(i + 1));    return result;    }    }    return Math.abs(m - n) == 1;    }}