[LeetCode] Find the Duplicate Number
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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
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.
思路1使用map存储元素及其对应索引。
class Solution { public: int findDuplicate(vector<int>& nums) { if (nums.empty()) return 0; unordered_map<int, int> m; for (int i = 0; i < nums.size(); ++i) { if (m.count(nums[i]) > 0) return nums[i]; else m[nums[i]] = i; } } };
思路2:双层遍历找出相同元素
class Solution { public: int findDuplicate(vector<int>& nums) { for (int i = 0; i < nums.size(); ++i) { for (int j = i+1; j < nums.size(); ++j) if (nums[i] == nums[j]) return nums[i]; } } };
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