[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]:

  1. 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;
}
}
04-14 13:41