[抄题]:
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
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
以为要用位运算,结果是:出现两次的value当作指针,数组形成环。所以用快慢指针
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[一句话思路]:
数组中的一步两步:value变成index,用数组包起来
一层/两层:
slow = nums[slow]; fast = nums[nums[fast]];
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 不知道为什么要从头再走一遍:直接找的中点是从中间断开的。为了从起点开始,需要再走一遍。
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
直接找的中点是从中间断开的。为了从起点开始,需要再走一遍。
[复杂度]:Time complexity: O(方) Space complexity: O(1)
[算法思想:迭代/递归/分治/贪心]:
[关键模板化代码]:
//start again and move to slow
fast = 0;
while (fast != slow) {
slow = nums[slow];
fast = nums[fast];
}
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
[代码风格] :
[是否头一次写此类driver funcion的代码] :
[潜台词] :
class Solution {
public int findDuplicate(int[] nums) {
//corner case
if (nums == null || nums.length == 0) return -1; //initialization: slow, fast
int slow = nums[0];
int fast = nums[nums[0]]; //find the number and store in slow
while (fast != slow) {
slow = nums[slow];
fast = nums[nums[fast]];
}
//store in slow now //start again and move to slow
fast = 0;
while (fast != slow) {
slow = nums[slow];
fast = nums[fast];
} //return
return fast;
}
}