Find the Duplicate Number

Nov 21, 2017


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


int findDuplicate1(vector<int>& nums) {
    vector<int> count(nums.size(),0);
    for(auto i:nums) {
        if (count[i]>1) return i;
    return -1;


int findDuplicate2(vector<int>& nums) {
    int t = 0;
    int n = nums.size() - 1;
    for(auto i:nums) {
        t ^= i;
    for(int i = 1;i <= n;i++)
        t ^= i;
    return t;


他是用了一个List Cycle中找环点的方式,这里的链表中的nxet就是用nums的值来表示的。

int findDuplicate(vector<int>& nums) {
    int n = nums.size();
    int slow = n;
    int fast = n;
    do {
        slow = nums[slow-1];
        fast = nums[nums[fast-1]-1];
    }while(slow != fast);
    slow = n;
    while(slow != fast) {
        slow = nums[slow - 1];
        fast = nums[fast - 1];
    return slow;

Hamming Distance

Perfect Squares

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