3.1 Distribute Candies
Alice has n candies, where the ith candy is of type candyType[i]. Alice noticed that she started to gain weight, so she visited a doctor.
The doctor advised Alice to only eat n / 2 of the candies she has (n is always even). Alice likes her candies very much, and she wants to eat the maximum number of different types of candies while still following the doctor's advice.
Given the integer array candyType of length n, return the maximum number of different types of candies she can eat if she only eats n / 2 of them.
Example 1:
Input: candyType = [1,1,2,2,3,3]
Output: 3
Explanation: Alice can only eat 6 / 2 = 3 candies. Since there are only 3 types, she can eat one of each type.
Example 2:
Input: candyType = [1,1,2,3]
Output: 2
Explanation: Alice can only eat 4 / 2 = 2 candies. Whether she eats types [1,2], [1,3], or [2,3], she still can only eat 2 different types.
Example 3:
Input: candyType = [6,6,6,6]
Output: 1
Explanation: Alice can only eat 4 / 2 = 2 candies. Even though she can eat 2 candies, she only has 1 type.
Constraints:
n == candyType.length
2 <= n <= 104
n is even.
-105 <= candyType[i] <= 105
秒了，题解如下，思路不复杂不多解释：
var distributeCandies = function(candyType) {
return Math.min(candyType.length/2,new Set(candyType).size)
};
3.2 Set Mismatch
You have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number.
You are given an integer array nums representing the data status of this set after the error.
Find the number that occurs twice and the number that is missing and return them in the form of an array.
Example 1:
Input: nums = [1,2,2,4]
Output: [2,3]
Example 2:
Input: nums = [1,1]
Output: [1,2]
Constraints:
2 <= nums.length <= 104
1 <= nums[i] <= 104
看到这道题一开始想法挺多的，但是二分查找似乎并不适用于这道题，于是写了个O(n)的解法，自我感觉有点僵硬：
var findErrorNums = function(nums) {
let ans = [];
let cnt = new Set();
for (let i of nums){
cnt.has(i)?ans.push(i):cnt.add(i);
}
for (let i=1;i<=nums.length;i++){
if (!cnt.has(i)) ans.push(i)
}
return ans
};
翻了题解，改写一个python的解法，大概思路是把结果数组中的[a,b]通过先凑出(a+b)/2和(b-a)/2然后算出a与b，一看就懂，可惜想不到：
var findErrorNums = function(nums) {
let n = nums.length;
let A = n*(n+1)/2, B = n*(n+1)*(2*n+1)/6;
for (let i of nums){
A -= i;
B -= i*i;
}
return [(B-A*A)/2/A, (B+A*A)/2/A]
};
3.3 Missing Number
Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
Example 4:
Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.
Constraints:
n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are unique.
题目问能不能写个空间复杂度O(1),时间复杂度O(n)的算法，这直接提示了我写出下面的方法，用个等差数列求和公式，做减法得出缺失值，不是最好的但优化空间也不大了，故不翻别人题解了：
var missingNumber = function(nums) {
let sum = (nums.length+1)*nums.length/2;
for (let i of nums){
sum -= i
}
return sum
};
