Missing Number

• Missing Number
  • 题目描述
  • 解法

Missing Number

题目描述

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?

解法

代码如下:

public static int missingNumber( int[] nums ) {
    int sum = 0, n = nums.length;
    for( int i = 0; i < n; ++i )
        sum += nums[i];
    return n * ( n + 1 ) / 2 - sum;
}

思考过程:

根据条件, 已知数组中丢失的元素是数值范围[0...n]中的某一个值, 那么用该范围所有元素的和减去丢失后所有元素的和就是丢失的值.

时空复杂度: 时间复杂度是O(n), 空间复杂度是O(1)