287. Find the Duplicate Number

标签：complex   ESS   sub   fas   and   ble   highlight   runner   detection   

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.

Example 1:

Input: [1,3,4,2,2]
Output: 2

Example 2:

Input: [3,1,3,4,2]
Output: 3

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.

Approach #1: Using hash table

time(O(N)) 　　space(O(n))

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int len = nums.size();
        vector<int> temp(len, 0);
        for (int i = 0; i < len; ++i) 
            if (++temp[nums[i]] == 2) return nums[i];
    }
};

Approach #2 Floyd‘s Tortoise and Hare (Cycle Detection)

time(O(n))　　space(O(1))

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int tortoise = nums[0];
        int hare = nums[0];
        // Find the intersection point of the two runner.
        do {
            tortoise = nums[tortoise];
            hare = nums[nums[hare]];
        } while (tortoise != hare);
        // Find the "entrance" to the cycle
        int ptr1 = nums[0];
        int ptr2 = tortoise;
        while (ptr1 != ptr2) {
            ptr1 = nums[ptr1];
            ptr2 = nums[ptr2];
        }
        return ptr1;
    }
};

Runtime: 8 ms, faster than 83.02% of C++ online submissions for Find the Duplicate Number.

287. Find the Duplicate Number

标签：complex   ESS   sub   fas   and   ble   highlight   runner   detection   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/9902351.html