【LeetCode】Single Number
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原文:
Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
解答:
常规解法:先对数组进行排序,然后通过按顺序判断每相邻两个数是否相同即可;
public class Solution { public int singleNumber(int[] A) { Arrays.sort(A); for(int i=0;i<=A.length-1;){ //注意数组只有一个的情况 if((i+1)<=A.length-1){ //判断A[i]是否是最后一个 if(A[i]!=A[i+1]) return A[i]; else i+=2; }else{ return A[i]; } } return -1; }}
位运算解法:使用异或运算 这个程序用了个小技巧:一个整数和它本身异或之后得到值是0,0与其他整数异或得到的是这个整数本身.
public int singleNumber(int[] A){if(A.length==0)return -1;int result=0;for(int i=0;i<A.length;i++){result^=A[i];}return result;}
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