Leet Code OJ 268. Missing Number [Difficulty: Medium]
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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?
翻译:
给定一个数组,包含n个不同的数,从0, 1, 2, …, n里面取出,找到那个缺失的数。
提示:你的程序应该运行在一个线性的时间复杂度上。你可以实现它仅仅使用常量的额外空间复杂度吗?
分析:
按照题意,数组只缺失了一个数,来构成数列0, 1, 2, …, n,由于数组可能是乱序的,如果按照传统方法遍历,将需要先做排序(增加时间复杂度),或者使用额外空间来记录(增加空间复杂度)。
由于只缺少了一个数,我们可以把未缺失的原数列的和计算出来,与数组的和对比,差值即缺失的数。
代码:
public class Solution { public int missingNumber(int[] nums) { int sum=0; for(int i=0;i<nums.length;i++){ sum+=nums[i]; } int rightSum=(nums.length+1)*(nums.length)/2; return rightSum-sum; }}
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