287 - Find the Duplicate Number

寻找重复数字

最新推荐文章于 2021-12-17 00:37:20 发布

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本文介绍了一个在不修改且只使用常数额外空间的情况下找到数组中唯一重复数字的算法。该算法利用了二分查找的思想，并通过迭代缩小搜索范围来降低时间复杂度。

Problem

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.

Code

class Solution {
public:
    int cmp(vector<int>& nums, int val, int& low, int& up) {
        int t = 0;
        for (vector<int>::iterator it = nums.begin(); it != nums.end(); ++it) {
            if (*it == val) {
                t++;
            } else if (*it > val) {
                up++;
            } else {
                low++;
            }
        }
        return t;
    }
    
    int findDuplicate(vector<int>& nums) {
        int n = nums.size() - 1;
        int l = 1, r = n;
        while (l < r) {
            int mid = (l + r) / 2, low = 0, up = 0;
            int cnt = cmp(nums, mid, low, up);
            if (cnt > 1) {
                return mid;
            }
            if (low >= mid) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
};