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题目:
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. For example, Given nums = [0, 1, 3] return 2. Note: Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
分析:
方法一:根据元素的顺序,进行异或^操作:
public int missingNumber(int[] nums) { int xor = 0, i = 0; for (i = 0; i < nums.length; i++) { xor = xor ^ i ^ nums[i]; } return xor ^ i; }
方法二:利用数学求和,用期望值减去实际值即可:
public int missingNumber(int[] nums) { int sum = 0; for(int i : nums) { sum += i; } int n = nums.length; int tmp = (n+1)*n/2; return tmp-sum; }
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原文地址:http://www.cnblogs.com/lasclocker/p/4986726.html