LeetCode 448. Find All Numbers Disappeared in an Array(查找数组中没有出现的数字)
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Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.
输入一个长度为n的数组(数组中存在重复的数字),查找[1,n]中没有出现的数字
Example:
Input:[4,3,2,7,8,2,3,1]Output:[5,6]
思路一:
首先利用sort()函数对数组进行排序;然后遍历数组,查找没有出现的数字;最后补充区间(数组中最大的数字,n]的数字。
class Solution {public: vector<int> findDisappearedNumbers(vector<int>& nums) { vector<int> disappearNums; sort(nums.begin(),nums.end()); int i=1; for(int n:nums){ while(n>=i){ if(n>i) disappearNums.push_back(i); i++; } } for(i+1;i<=nums.size();i++) disappearNums.push_back(i); return disappearNums; }};
思路二:
直接遍历数组,将元素值位置m=(abs(nums[i])-1)对应的元素取反(即nums[m] = -nums[m]),以标志该位置数(该位置数+1,因为index从0开始)已经出现过;重新遍历数组,数值为正的数即为没有出现过的数字。class Solution {public: vector<int> findDisappearedNumbers(vector<int>& nums) { int len = nums.size(); for(int i=0; i<len; i++) { int m = abs(nums[i])-1; // index start from 0 nums[m] = nums[m]>0 ? -nums[m] : nums[m]; } vector<int> res; for(int i = 0; i<len; i++) { if(nums[i] > 0) res.push_back(i+1); } return res; }};
思路三:
使用unique()+nums.erase()函数对数组去重;然后使用insert()函数对数组进行重建,即位置i对应的元素为i+1;发生插入操作的位置,插入的元素即为没有出现的元素。阅读全文
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