260. Single Number III

题目：

Given an array of numbers

nums

, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given

nums = [1, 2, 1, 3, 2, 5]

, return

[3, 5]

.

Note:

The order of the result is not important. So in the above example,

[5, 3]

is also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

链接： http://leetcode.com/problems/single-number-iii/

题解：

昨晚卡到现在。但昨晚睡觉前google了一下，看到了几篇文章以后才理解。还是需要加强bit manipulation。 归根结底是大学时数字电路没学好 -_____-!! 我不堪的本科生涯...略去一亿字。

这里我们主要是先异或所有数字，得到 a^b， 因为a和b不相等，所以 a^b里至少有一个bit位k，k = 1。 然后我们就可以用这个第k位来进行判断， 因为a和b里只有一个数第k位 = 1，我们对输入数组所有第k位为1的数字进行异或，就可以找到第k位为1的仅出现一次的数字， 再用这个数字a和 a^b进行异或，我们也就求得了b。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
public int[] singleNumber(int[] nums) {
int[] res = {-1, -1};
if(nums == null || nums.length == 0) {
return res;
}

int a_XOR_b = 0;
for(int i = 0; i < nums.length; i++) {      // find a ^ b
a_XOR_b ^= nums[i];
}
int k = 0;                              // find one bit in a^b == 1
for(int i = 0; i < 31; i++) {           // that means only one num in a, b has 1 on kth bit
if(((a_XOR_b >> i) & 1) == 1) {
k = i;
break;
}
}
int a = 0;
for(int i = 0; i < nums.length; i++) {   // since only one num in a, b has 1 on kth bit
if(((nums[i] >> k) & 1) == 1) {      // we can XOR all nums has 1 on kth bit
a ^= nums[i];                    // duplicate nums will be evened out
}
}
int b = a ^ a_XOR_b;
res[0] = a;
res[1] = b;
return res;
}
}

二刷：

也是用一刷一样的方法，比较tricky

Java:

public class Solution {
public int[] singleNumber(int[] nums) {
int[] res = {-1, -1};
if (nums == null || nums.length == 0) {
return res;
}
int a_XOR_b = 0;
for (int i : nums) {
a_XOR_b ^= i;
}
int k = 0;
for (int i = 0; i < 32; i++) {
if (((a_XOR_b >> i) & 1) == 1) {
k = i;
break;
}
}
res[0] = 0;
for (int i : nums) {
if (((i >> k) & 1) == 1) {
res[0] ^= i;
}
}
res[1] = a_XOR_b ^ res[0];
return res;
}
}

题外话：

其实本科时，学校开设的课程挺不错的，老师也努力务实，都怪自己没把心思放在学习上。虽然专业是EE，但像什么基数排序，霍夫曼编码，游程码，最短路径之类的东西，老师们也都多少介绍过。信息量很大，作业和考试也难。真要四年扎扎实实学下来的话，也不至于现在一把年纪了还在刷题。 过去的都过去了，接下来好好努力吧。

三刷:

这次就比较熟练了，尝试一些不同的编程风格。 4/4/2016。

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
public int[] singleNumber(int[] nums) {
int[] res = new int[2];
if (nums == null || nums.length < 2) return res;
int a_xor_b = 0;
for (int num : nums) { a_xor_b ^= num; }
int diffIndex = 0;
while (diffIndex < 32) {
if (((a_xor_b >> diffIndex) & 1) == 1) { break; }
diffIndex++;
}
int num1 = 0;
for (int num : nums) {
if (((num >> diffIndex) & 1) == 1) { num1 ^= num; }
}
res[0] = num1;
res[1] = a_xor_b ^ num1;
return res;
}
}

Reference:
https://groups.google.com/forum/#!topic/pongba/9KCA7b484OE https://groups.google.com/forum/#!topic/pongba/drV6dytcoJE http://www.matrix67.com/blog/archives/511