"One Day together Leetcode" #260. Single number III

One Day together Leetcode

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(i) Title

Given An array of numbers nums, in which exactly, elements appear only once and all the other elements appear exactly t Wice. Find >the, elements, 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. The above example, [5, 3] is also correct.
+ Your algorithm should run in linear runtime complexity.
+ Could Implement it using only constant space complexity?

(ii) Problem solving

The main idea: there are two numbers in an array only once, the other number appears two times, find out the two numbers

Problem-solving ideas: A comparison of "one Day together Leetcode" #137. Single number, if there are two occurrences only once, then use the XOR operation to find out the XOR value of the two numbers

From this XOR value, these two numbers are different, then you can find the different bits, take a different bit, such as the first M bit.

The whole array is divided into two categories, one for the M-bit is 1, the class m is 0, so that in each group number only a single occurrence of the number, it is easy to use the XOR operation to solve.
See the code for specific ideas:

classSolution { Public: vector<int>Singlenumber ( vector<int>& Nums) { vector<int>RetintSize = Nums.size ();if(Size = =0)returnRetinttemp = nums[0]; for(inti =1; i < size; i++) temp^=nums[i];//Find the final value of all or after        unsigned intBit for(inti =0; I < +; i++) {//Find the first 1 bit in tempbit=1<<i;if((temp&bit) ==bit) Break; }intTemp1 =0, Temp2 =0;//TEMP1 is the first number that appears only once, Temp2 is the second         for(inti =0; i < size; i++) {if((nums[i]&bit) = =0) {//Group, 0 on the first m bitTemp1^=nums[i]; }Else{Temp2^=nums[i];//Group 1 on the M-bit}} ret.push_back (TEMP1); Ret.push_back (TEMP2);returnRet }};

Related Topics:
"One Day together Leetcode" #137. Single number II
"One Day together Leetcode" #137. Single number

"One Day together Leetcode" #260. Single number III