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

    Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one.

    Note:

    1. You must not modify the array (assume the array is read only).
    2. You must use only constant, O(1) extra space.
    3. Your runtime complexity should be less than O(n2).
    4. There is only one duplicate number in the array, but it could be repeated more than once.

    Credits:
    Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

    Analysis:

    if duplicated number is less than X, then there must be more than X numbers in the array less than X.

    So, we use binary search from 1 to n, count the number of number <= mid, and determine whether is the duplicated number.

    Solution:

    public class Solution {
        public int findDuplicate(int[] nums) {
            if (nums.length == 0)
                return 0;
    
            int start = 1, end = nums.length - 1; // start is 1, end is n.
            while (start <= end) {
                int mid = start + (end - start) / 2;
                // count the numbers no larger than mid.
                int count = 0;
                for (int num : nums)
                    if (num <= mid) {
                        count++;
                    }
    
                if (count > mid) {
                    end = mid - 1;
                } else {
                    start = mid + 1;
                }
            }
            return start;
        }
    }
  • 相关阅读:
    NSFileHandle学习
    NSFileManager学习创建拷贝删除
    电量电费的尖峰平谷
    电表的付费模式
    电能表类型
    oracle递归
    相关子查询和非相关子查询
    再说exists 关键字,和inner join 差别大
    发电厂和电力局的关系
    js的toFixed()方法
  • 原文地址:https://www.cnblogs.com/lishiblog/p/5836842.html
Copyright © 2011-2022 走看看