287 Find the Duplicate Number
1. Question
Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(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(n^2)
.There is only one duplicate number in the array, but it could be repeated more than once.
2. Implementation
(1) Binary Search
class Solution {
public int findDuplicate(int[] nums) {
int start = 0, end = nums.length - 1, mid = 0;
while (start + 1 < end) {
mid = start + (end - start) / 2;
int count = countNum(nums, mid);
if (count <= mid) {
start = mid + 1;
}
else {
end = mid;
}
}
return countNum(nums, start) > start ? start : end;
}
public int countNum(int[] nums, int target) {
int count = 0;
for (int num : nums) {
if (num <= target) {
++count;
}
}
return count;
}
}
(2) Two Pointers
public class Solution {
public int findDuplicate(int[] nums) {
// Similar to Linked List Cycle II
// Use Two Pointers to find the begnning of loop
// This is where the duplicate occur
int fast = 0, slow = 0;
do {
fast = nums[nums[fast]];
slow = nums[slow];
} while (fast != slow);
slow = 0;
while (fast != slow) {
fast = nums[fast];
slow = nums[slow];
}
return slow;
}
}
3. Time & Space Complexity
Binary Search: 时间复杂度O(nlogn), 空间复杂度O(1)
Two Pointers: 时间复杂度O(n), 空间复杂度O(1)
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