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pu406981650
V2EX    Java

lintcode 中 tle 错误,求大佬帮忙看下

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  •   pu406981650 Sep 10, 2017 2641 views
    This topic created in 3151 days ago, the information mentioned may be changed or developed.

    题目: Strings Homomorphism

    public class Solution { /** * @param s a string * @param t a string * @return true if the characters in s can be replaced to get t or false */ public boolean isIsomorphic(String s, String t) { // Write your code here if(s == null && t == null) return true; else if(s == null || t == null) return false; else if(s.length() != t.length()) return false; else{ String s1 = ""; String s2 = ""; int j = 0; for(int i = 0; i < s.length(); i++){ int dis1 = s.indexOf(s.charAt(i)); if(i == dis1){ s1 += j; j++; } else{ s1 += s1.charAt(dis1); } } int l = 0; for(int k = 0; k < t.length(); k++){ int dis2 = t.indexOf(t.charAt(k)); if(k == dis2){ s2 += l; l++; } else{ s2 += s2.charAt(dis2); } } boolean flag = s1.equals(s2); return flag; } } } 

    还有另外一个 题目: First Position Unique Character

    public class Solution { /* * @param s: a string * @return: it's index */ public int firstUniqChar(String s) { // write your code here if (s == null) { return -1; } boolean flag = false; int i = 0; for (; i < s.length() - 1; i++) { String myStr = s.substring(0, i) + s.substring(i + 1); String ch = Character.toString(s.charAt(i)); if (myStr.contains(ch)) { continue; } else { flag = true; break; } } if (flag) { return i; } else { return -1; } } } 
    2 replies    2017-09-11 08:44:09 +08:00
    LxExExl
        1
    LxExExl  
       Sep 11, 2017 via iPhone
    第一题不需要 for 循环 长度一样就 true 了吧
    第二题 naive 的做法也应该是弄个 hashset O(N) 吧
    RecursiveG
        2
    RecursiveG  
       Sep 11, 2017
    给楼主一个 Strings Homomorphism 的测试样例,
    然后建议楼主自己研究下为啥不对。
    new Solution().isIsomorphic("abcdefghijkla","abcdefghijkll")
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