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287. Find the Duplicate Number
Description:
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.
Difficulty:Medium
Example:
Input: [1,3,4,2,2]
Output: 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
.
4.There is only one duplicate number in the array, but it could be repeated more than once.
方法1:排序、Hash、取负标记,不符合题意
思路:
很容易联想到题目:448.Find All Numbers Disappeared in an Array
利用改变元素正负号和取绝对值的方法,标记是否出现过,如果出现过,直接返回index
,但是不符合题意,改变了原数组。
class Solution {
public:
int findDuplicate(vector<int>& nums) {
for (int i = 0; i < nums.size(); i++) {
int index = abs(nums[i]) - 1;
if (nums[index] < 0)
return index + 1;
else
nums[index] = -nums[index];
}
}
};
方法2:快慢指针
- Time complexity :
- Space complexity :
思路:
尝试连接成环,肯定能够连接成以重复元素作为入口的环。
接下来问题就变成了找环的入口,经典公式,第一次快慢指针相遇时a+x=(a+b+x)/2 => a+x=b
,然后快指针归0,并且变成慢指针速度,快慢指针同时走,第二次相遇便是入口。
关键点:
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int slow = nums[0];
int fast = nums[nums[0]];
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
fast = 0;
while (slow != fast) {
slow = nums[slow];
fast = nums[fast];
}
return slow;
}
};