leetCode No.268 Missing Number
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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?
标签:Array、Math、Bit Manipulation
相似问题:(H) First Missing Positive,(E) Single Number,(H) Find the Duplicate Number
题意
给定一个整数n和一个数组,数组元素为1,2,3,4…n丢掉一个,找到丢掉的元素。
解题思路
根据长度n求得差值为1的等差数列的和,即为0到n所有数的和。再循环数组,用和减去数组中的元素即可得到缺失的元素。
代码
public class Solution { public int missingNumber(int[] nums) { int a = nums[0]; int index = 0; for (int i = 0;i < nums.length;i++) { if (nums[i] == a + i) { if (i == nums.length - 1) { index = nums[i] + 1; } } else { index = a + i; break; } } return index; }}
相关链接
源代码(github)
原题
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