217. Contains Duplicate

题目描述

Given an array of integers, find if the array contains any duplicates.

Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct.

Example 1:

Input: [1,2,3,1]
Output: true
Example 2:

Input: [1,2,3,4]
Output: false
Example 3:

Input: [1,1,1,3,3,4,3,2,4,2]
Output: true

Qiang的思路

class Solution:
    # def heapify(self, nums, i):
    #     if i>(len(nums)-1-1)//2:
    #         return
    #     if i*2+2>=len(nums):
    #         if nums[i]<nums[i*2+1]:
    #             nums[i], nums[i*2+1] = nums[i*2+1], nums[i]
    #     else:
    #         if nums[i]<nums[i*2+1] and nums[i*2+1]>nums[i*2+2]:
    #             nums[i], nums[i*2+1] = nums[i*2+1], nums[i]
    #             self.heapify(nums, i*2+1)
    #         elif nums[i]<nums[i*2+2]:
    #             nums[i], nums[i*2+2] = nums[i*2+2], nums[i]
    #             self.heapify(nums, i*2+2)
        
    def containsDuplicate(self, nums: List[int]) -> bool:
        # for i in range((len(nums)-1-1)//2, -1, -1):
        #     self.heapify(nums, i)
        # for i in range((len(nums)-1-1)//2+1):
        #     if nums[i] == nums[i*2+1] or (i*2+2<=len(nums)-1 and nums[i] == nums[i*2+2]):
        #         return True
        # return False
        return len(nums) != len(set(nums))

class Solution {
public:
    bool containsDuplicate(vector<int>& nums) {
        set<int> s;
        for(int i=0; i<nums.size(); i++)
            s.insert(nums[i]);
        return s.size() != nums.size();
    }
};