LeetCode-Easy部分中标签为Math 268. Missing Number
2017-04-05 08:56
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原题
Given an array containing n distinct numbers taken from0, 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?
题目分析
找出序列中缺少的那个数,例如,nums=[0,1,2],因为必然缺少一个元素,所以缺少为3;nums=[1,2,3],因为元素从0开始,所以缺少的数为0。这个题最简单的解法,位操作运算,异或。我们知道 a^b^b = a,两个相同的数异或后都被抵消。借助这点,我们应用到此题上,索引和元素都出现的,抵消后,剩余的那个缺失元素,必然仅索引出现一次的了,
代码实现
public class Solution { public int MissingNumber(int[] nums) { int xor = 0, i = 0; for (i = 0; i <nums.Length; i++) { xor = xor ^ i ^ nums[i]; } //最后再和个数i异或,这样所有的索引与下标都异或了,找到缺失元素。 return xor^i; } }
更多标签为Math的题目
http://blog.csdn.net/daigualu/article/details/69101448
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