題目來源
找一個數(shù)組中重復(fù)的數(shù)字，size=n+1，數(shù)字范圍1-n，不能用額外空間，不能改變原數(shù)組，時間復(fù)雜度小于O(N^2)，想了半天，只能用二分。
代碼如下：
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int n = nums.size();
int l = 1, r = n - 1;
while (l < r) {
int mid = (l + r) / 2;
int smallOrEquals = 0;
for (int i=0; i<n; i++)
if (nums[i] <= mid)
smallOrEquals++;
if (smallOrEquals > mid)
r = mid;
else
l = mid + 1;
}
return l;
}
};
還有O(N)的做法…完全想不到，類似于之前鏈表找循環(huán)的起始點，代碼如下：
class Solution {
public:
int findDuplicate(vector<int>& nums) {
int slow = nums[0], fast = nums[nums[0]];
while (slow != fast) {
slow = nums[slow];
fast = nums[nums[fast]];
}
int slow2 = 0;
while (slow != slow2) {
slow = nums[slow];
slow2 = nums[slow2];
}
return slow;
}
};
