zoukankan      html  css  js  c++  java
  • [LeetCode] Find the Duplicate Number

    There are mainly two solutions to solve this problem.

    The first one is very clever, using the idea of cycle detection, and runs in O(n) time. You may refer to this post, whose code is rewritten below.

     1 class Solution {
     2 public:
     3     int findDuplicate(vector<int>& nums) {
     4         int n = nums.size(), slow = n - 1, fast = n - 1;
     5         do {
     6             slow = nums[slow] - 1;
     7             fast = nums[nums[fast] - 1] - 1;
     8         } while (slow != fast);
     9         fast = n - 1; 
    10         do {
    11             slow = nums[slow] - 1;
    12             fast = nums[fast] - 1;
    13         } while (slow != fast);
    14         return slow + 1;
    15     }
    16 };

    The second one uses binary search to search for possible numbers in ranges [1, n] and eliminate one half at each time. You may refer to this nice post for a good explanation. The code is rewritten below in C++.

     1 class Solution {
     2 public:
     3     int findDuplicate(vector<int>& nums) {
     4         int n = nums.size(), l = 1, r = n - 1;
     5         while (l < r) {
     6             int m = l + ((r - l) >> 1);
     7             int cnt = notGreaterThan(nums, m);
     8             if (cnt <= m) l = m + 1;
     9             else r = m;
    10         } 
    11         return l;
    12     }
    13 private:
    14     int notGreaterThan(vector<int>& nums, int target) {
    15         int cnt = 0;
    16         for (int num : nums)
    17             if (num <= target)
    18                 cnt++;
    19         return cnt;
    20     }
    21 };
  • 相关阅读:
    ubuntu安装后要做什么
    JavaScript错误处理
    jQuery 事件
    display的用法
    百度排名的原理
    什么是ajax?
    CSS文档流
    引用CSS的方法
    jQuery的安装方式
    禁止WPS2019开机自启动
  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4844534.html
Copyright © 2011-2022 走看看