标签：example new time ica dup solution method nts and
Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements that appear twice in this array.
Could you do it without extra space and in O(n) runtime?
Example:
Input:
[4,3,2,7,8,2,3,1]
Output:
[2,3]
idea: similar methods with 448 Find All Numbers Disappeared in an Array
Solution 1: If nums[index] is already <0, index+1 is the duplicated number
1 class Solution { 2 public: 3 vector<int> findDuplicates(vector<int>& nums) { 4 int n=nums.size(); 5 vector<int> res; 6 for (int i=0;i<n;i++){ 7 int index=abs(nums[i])-1; 8 if (nums[index]<0) res.push_back(abs(nums[i])); //diff with 448 9 else nums[index]=-nums[index]; 10 } 11 return res; 12 } 13 };
Solution 2: If nums[i]!= i+1, nums[i] is the duplicated number.
1 class Solution { 2 public: 3 vector<int> findDuplicates(vector<int>& nums) { 4 int n=nums.size(); 5 vector<int> res; 6 for (int i=0;i<n;i++){ 7 int index=nums[i]-1; 8 if (nums[index]!=nums[i]) { 9 swap (nums[i],nums[index]); 10 --i; 11 } 12 } 13 for (int i=0;i<n;i++){ 14 if (nums[i]!=i+1) res.push_back(nums[i]); //diff with 448 15 } 16 return res; 17 } 18 };
Solution 3: If nums[i]>2*n, i+1 is the duplicated number.
1 class Solution { 2 public: 3 vector<int> findDuplicates(vector<int>& nums) { 4 int n=nums.size(); 5 vector<int> res; 6 for (int i=0;i<n;i++){ 7 nums[(nums[i]-1)%n]+=n; 8 } 9 for (int i=0;i<n;i++){ 10 if (nums[i]>2*n) res.push_back(i+1); //diff with 448 11 } 12 return res; 13 } 14 };
442. Find All Duplicates in an Array
标签：example new time ica dup solution method nts and
原文地址：http://www.cnblogs.com/anghostcici/p/6664352.html
