Find the Duplicate Number
題意：
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.
Note:
- You must not modify the array (assume the array is read only).
- You must use only constant, O(1) extra space.
- Your runtime complexity should be less than O(n2).
- There is only one duplicate number in the array, but it could be repeated more than once.
解題思路：
O(N^2) 解法，程式碼如下：
public class Solution {
public int findDuplicate(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (nums[i] == nums[j]) {
return nums[i];
}
}
}
return 0;
}
}
有網友提供了一個類似linked list用快慢指針找cycle的方式，Floyd Cycle Detection，只需要走訪兩遍陣列即可，複雜度為O(N)，其程式碼如下
使用這個方式的條件是有n+1個數，且值都在1到n之間，否則會有index overflow的問題。
public class Solution {
public int findDuplicate(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int slow = 0;
int fast = 0;
int finder = 0;
while (true) {
slow = nums[slow];
fast = nums[nums[fast]];
if (slow == fast) {
break;
}
}
while (true) {
slow = nums[slow];
finder = nums[finder];
if (slow == finder) {
return slow;
}
}
}
}