标签：while www. splay Once 空间 else val += c++
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:
寻找一个数组里的重复数，由于只能O(1)的空间复杂度，所以哈希表之类的就不能用了。
解法1: 双指针，寻找环。类似于 142. Linked List Cycle II
Treat each (key, value) pair of the array as the (pointer, next) node of the linked list, thus the duplicated number will be the begin of the cycle in the linked list. Besides, there is always a cycle in the linked list which starts from the first element of the array.
解法2: Binary Search
Java:
public int findDuplicate(int[] nums) {
int slow = 0;
int fast = 0;
do{
slow = nums[slow];
fast = nums[nums[fast]];
} while(slow != fast);
int find = 0;
while(find != slow){
slow = nums[slow];
find = nums[find];
}
return find;
}
Java:
public int findDuplicate(int[] nums) {
int l = 1,r = nums.length - 1;
while(l < r){
int m = (l + r) / 2;
int c = 0;
for(int i: nums){
if(i <= m){
c++;
}
}
//if c < m,
if(c > m){
r = m;
}else{
l = m + 1;
}
}
return r;
}
Python:
# Two pointers method
class Solution(object):
def findDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
slow = nums[0]
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
Python:
# Time: O(nlogn)
# Space: O(1)
# Binary search method.
class Solution(object):
def findDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
left, right = 1, len(nums) - 1
while left <= right:
mid = left + (right - left) / 2
# Get count of num <= mid.
count = 0
for num in nums:
if num <= mid:
count += 1
if count > mid:
right = mid - 1
else:
left = mid + 1
return left
Python:
# Time: O(n)
# Space: O(n)
class Solution(object):
def findDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
duplicate = 0
# Mark the value as visited by negative.
for num in nums:
if nums[abs(num) - 1] > 0:
nums[abs(num) - 1] *= -1
else:
duplicate = abs(num)
break
# Rollback the value.
for num in nums:
if nums[abs(num) - 1] < 0:
nums[abs(num) - 1] *= -1
else:
break
return duplicate
C++:
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int low = 1, high = nums.size() - 1;
while (low < high) {
int mid = low + (high - low) * 0.5;
int cnt = 0;
for (auto a : nums) {
if (a <= mid) ++cnt;
}
if (cnt <= mid) low = mid + 1;
else high = mid;
}
return low;
}
};
C++:
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int slow = 0, fast = 0, t = 0;
while (true) {
slow = nums[slow];
fast = nums[nums[fast]];
if (slow == fast) break;
}
while (true) {
slow = nums[slow];
t = nums[t];
if (slow == t) break;
}
return slow;
}
};
类似题目：
[LeetCode] 142. Linked List Cycle II 链表中的环 II
[LeetCode] 287. Find the Duplicate Number 寻找重复数
标签：while www. splay Once 空间 else val += c++
原文地址：https://www.cnblogs.com/lightwindy/p/9602295.html
