Leetcode刷题小结(1)——异或

该篇博客总结了leetcode上几个关于bit manipulation（异或）的题。389题是字符间的异或，136和268题是关于数的异或。

389. Find the Difference

Given two strings s and t which consist of only lowercase letters. String t is generated by random shuffling string s and then add one more letter at a random position. Find the letter that was added in t.

Example:

Input:
s = "abcd"
t = "abcde"

Output:e
Explanation:'e' is the letter that was added.

class Solution {
    public char findTheDifference(String s, String t) {
        //将s和t的所有字符异或，得到的结果就是新加的字符  类似于136题寻找只出现1次的数字
    	char res=0;//将ASCII码为0的字符给c
    	for(char c:s.toCharArray()) {
    		res^=c;
    	}
    	for(char c:t.toCharArray()) {
    		res^=c;
    	}
		return res;
    }
}

136. Single Number

Given a non-empty 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?

Example 1:

Input: [2,2,1]
Output: 1

Example 2:

Input: [4,1,2,1,2]
Output: 4

class Solution {
    public int singleNumber(int[] nums) {
        //异或运算 A^B^A=B, A^0=A, A^B^C^B=A^(B^C^B)=A^C
    	int res=0;
        for(int i=0;i

268. Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1:

Input: [3,0,1]
Output: 2

Example 2:

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

class Solution {
    public int missingNumber(int[] nums) {
        int missing = nums.length;
        for (int i = 0; i < nums.length; i++) {
            missing ^= i ^ nums[i]; //下标对应的值与下标异或
        }
        return missing;
    }
}