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https://oj.leetcode.com/problems/single-number/
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?
public class Solution { public int singleNumber(int[] A) { int number = 0; HashMap<Integer, Integer> map = new HashMap(); for (int i = 0; i < A.length; i++) { if (map.get(A[i]) == null || map.get(A[i]) == 0) { map.put(A[i], 1); } else if (map.get(A[i]) == 1) { map.put(A[i], 2); } } Iterator<Integer> it = map.keySet().iterator(); while (it.hasNext()) { Integer keyString = it.next(); if (map.get(keyString) == 1) return keyString; } return number; } }
1. 最复杂的方法是，两次遍历，时间复杂度O(n^2)。
2. 上面的解法是，放入map，每次放入时候判断get是不是已经有了，有了就值为2。最后取出value为1的。时间复杂度为O(n)。
3. 还有一个很tricky的解法。用异或的方法。思路就是每位bit出现2次就清零，所以可以不断异或运算得出最终结果。
public class Solution { public int singleNumber(int[] A) { int result = 0; for(int i = 0; i < A.length; i++){ result = result ^ A[i]; } return result; } }
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原文地址：http://www.cnblogs.com/NickyYe/p/4221067.html
