Two Strings Are Anagrams
判断两个字符串是否为重组的,没想到太好的方法,直接用计数法简单粗暴。
public class Solution { /** * @param s: The first string * @param b: The second string * @return true or false */ public boolean anagram(String s, String t) { // write your code here int[] num = new int[270]; char[] s1 = s.toCharArray(); char[] t1 = t.toCharArray(); if (s.length() != t.length()) return false; for (int i = 0; i < s.length(); ++i) { num[s1[i]]++; --num[t1[i]]; } for (int i = 0; i < num.length; ++i) { if (num[i] != 0) return false; } return true; } };
标准答案:
public class Solution { /** * @param s: The first string * @param b: The second string * @return true or false */ public boolean anagram(String s, String t) { if (s.length() != t.length()) { return false; } int[] count = new int[256]; for (int i = 0; i < s.length(); i++) { count[(int) s.charAt(i)]++; } for (int i = 0; i < t.length(); i++) { count[(int) t.charAt(i)]--; if (count[(int) t.charAt(i)] < 0) { return false; } } return true; } };
Compare Strings
public class Solution { /** * @param A : A string includes Upper Case letters * @param B : A string includes Upper Case letter * @return : if string A contains all of the characters in B return true else return false */ public boolean compareStrings(String A, String B) { // write your code here int[] num = new int[256]; for (int i = 0; i < A.length(); ++i) { ++num[(int) A.charAt(i)]; } for (int i = 0; i < B.length(); ++i) { --num[(int)B.charAt(i)]; if (num[(int)B.charAt(i)] < 0) return false; } return true; } }
时间: 2024-10-16 04:03:28