leetCode No.268 Missing Number

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