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LeetCode2.1.23~2.1.24(Single Number与Single Number II)

2015-08-22 17:16 459 查看

这又是两道利用逻辑运算和位运算来得出结果的题目代码如下

2.1.23 Single Number

描述

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?

其实这种思路不仅仅用来处理出现两次的情况，它实际上可以找出任何一个出现奇数次的数据

public static int solution2_1_23(int[] array){
int x=0;
for(int a:array){
x=x^a;
}
return x;
}

2.1.24 Single Number II

描述

Given an array of integers, every element appears three times except for one. Find that single one.

Note: Your algorithm should have a linear runtime complexity. Could you implement it without using

extra memory?

public static int solution2_1_24(int[] array){
int[] bitnum=new int[32];
int res=0;
for(int i=0;i<32;i++){
for(int j=0;j<array.length;j++){
bitnum[i]+=(array[j]>>i)&1;
}
int temp=0;
if(bitnum[i]%3!=0){
temp=1;
}
else
temp=0;
res=res|(temp<<i);
}
return res;
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