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?
[Thought]
关于single number的解法,因为只有一个未知数,比较直观:http://fisherlei.blogspot.com/2013/11/leetcode-single-number-solution.html。
这题有两个未知数,直接做异或肯定是不行的,那么如何通过一些变换把这道题分解开,使得可以应用Single Number的解法来做,才是这个题目有意思的地方。
首先,对于数组A, 假设存在b,c两个数字,在数组中只出现了一次,那么对于整个数组进行异或操作的话,
^[A] = b^c , 因为其他的数因为出现了两次,异或的过程中就被清零了。
但是仅仅通过最后异或出来的值,是没办法求出b和c的值的,但是足以帮我们把b和c划分到不同的子数组中去。
一个整数有32位bit,对于b和c,除非两者是相同的数,否则一定存在第K位bit,两者是不同的。看下面的例子,
当找到这个K以后,就可以按照第K位bit是否等于1,将A数组划分成两个子数组,而这两个子数组分别包含了b和c,那么剩下的就只需要把single number的算法直接应用到这两个子数组上,就可以得到b和c了。
[Code]
1: class Solution {
2: public:
3: vector<int> singleNumber(vector<int>& nums) {
4: int length = nums.size();
5: // get the xor result of the array, b ^ c
6: int xor_result = 0;
7: for(int i =0; i< length; i++) {
8: xor_result ^= nums[i];
9: }
10: // get the K of first bit, which equals 1
11: int first_one_index = 0;
12: for(first_one_index =0; first_one_index< 32; first_one_index++) {
13: if((xor_result>>first_one_index) & 1 == 1) {
14: break;
15: }
16: }
17: // use k to split the array into two part
18: // xor the sub array, if the element's Kth bit also equals 1, b
19: int xor_twice = 0;
20: for(int i =0; i< length; i++) {
21: if((nums[i]>>first_one_index) & 1 == 1) {
22: xor_twice ^= nums[i];
23: }
24: }
25: // with b, easy to get c by math
26: vector<int> result = {xor_twice, xor_result ^ xor_twice };
27: return result;
28: }
29: };
Git hub: https://github.com/codingtmd/leetcode/blob/master/src/Single_Number_III.cpp
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