• 268. Missing Number


    Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

    Example 1

    Input: [3,0,1]
    Output: 2
    

    Example 2

    Input: [9,6,4,2,3,5,7,0,1]
    Output: 8
    

    Note:
    Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

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

    解题思路:

    在最后添加一个-1值,对数组从0-N对每个数字进行归位,最后数字不符合的就是丢失的那个数字。

    1. class Solution {  
    2. public:  
    3.     int missingNumber(vector<int>& nums) {  
    4.         if(nums.size()==0) return 0;  
    5.         nums.push_back(-1);  
    6.         for(int i=0;i<nums.size();i++){  
    7.             int temp;  
    8.             int index=nums[i];  
    9.             if(index==-1) break;  
    10.               
    11.             while(1){  
    12.                 temp = nums[index];  
    13.                 nums[index] = index;  
    14.                 if(temp == nums[index] || temp ==-1) break;  
    15.                 index = temp;  
    16.             }  
    17.               
    18.         }  
    19.          for(int i=0;i<=nums.size();i++)  
    20.           if(nums[i]!=i) return i;  
    21.           
    22.     }  
    23. };  
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  • 原文地址:https://www.cnblogs.com/liangyc/p/8847510.html
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