Write a javascript function,Given an array of integers, find the one that appears an odd number of times. There will always be only one integer that appears an odd number of times. Examples [7] should return 7, because it occurs 1 time (which is odd). [0] should return 0, because it occurs 1 time (which is odd). [1,1,2] should return 2, because it occurs 1 time (which is odd). [0,1,0,1,0] should return 0, because it occurs 3 times (which is odd). [1,2,2,3,3,3,4,3,3,3,2,2,1] should return 4, because it appears 1 time (which is odd).

Sure, here's a JavaScript function that solves the problem you described with the given examples:

function findOdd(arr) {
  let count = {};
  for (let i = 0; i < arr.length; i++) {
    if (!count[arr[i]]) {
      count[arr[i]] = 1;
    } else {
      count[arr[i]]++;
    }
  }
  for (let key in count) {
    if (count[key] % 2 !== 0) {
      return parseInt(key);
    }
  }
}

This function uses an object to store the count of each integer in the array. We then loop through the object and return the integer that appears an odd number of times.

相关问题

Write a javascript functionGiven an array of integers, find the one that appears an odd number of times. There will always be only one integer that appears an odd number of times.

Sure, here's a JavaScript function that solves the problem you described:

function findOdd(arr) {
  let result = 0;
  for (let i = 0; i < arr.length; i++) {
    result ^= arr[i];
  }
  return result;
}

This function uses the XOR (^) operator to find the integer that appears an odd number of times. The XOR operator returns a 1 in each bit position for which the corresponding bits of either but not both operands are 1s. By XORing all the integers in the array together, we end up with the integer that appears an odd number of times.

For an array of integers [a1,a2,…,an], let's call the value |a1−a2|+|a2−a3|+⋯+|an−1−an| the contrast of the array. Note that the contrast of an array of size 1 is equal to 0 . You are given an array of integers a . Your task is to build an array of b in such a way that all the following conditions are met: b is not empty, i.e there is at least one element; b is a subsequence of a, i.e b can be produced by deleting some elements from a (maybe zero); the contrast of b is equal to the contrast of a . What is the minimum possible size of the array b ? Input The first line contains a single integer t (1≤t≤104 ) — the number of test cases. The first line of each test case contains a single integer n (1≤n≤3⋅105) — the size of the array a . The second line contains n integers a1,a2,⋅,an (0≤ai≤109 ) — elements of the array itself. The sum of n over all test cases doesn't exceed 3⋅105.

题意简述：

给定一个长度为 $n$ 的整数数组 $a$，求该数组的一个非空子序列 $b$，使得 $b$ 的对比度与 $a$ 的对比度相等。对比度的定义为 $|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n-1} - a_n|$。

解题思路：

根据对比度的定义，我们可以得到：

$|a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n-1} - a_n| = |a_1 - a_n| + |a_n - a_{n-1}| + \cdots + |a_2 - a_1|$

我们可以令 $b$ 为 $a$ 中某个值最大/最小的数，然后在 $a$ 中寻找与 $b$ 相邻的两个数，即 $a_i$ 和 $a_j$，满足 $a_i < b < a_j$ 或 $a_i > b > a_j$，然后将这个区间的数取出来组成子序列 $b$ 即可。

为什么这样做是正确的呢？因为对于一个区间 $[a_i, a_j]$，我们可以将 $b$ 看作是左右两个端点中较大/较小的那个，那么我们只需要考虑 $a_i$ 和 $a_j$ 与 $b$ 的关系即可。

时间复杂度为 $O(n)$。

参考代码：