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  • [LeetCode] Find the Duplicate Number

     

    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:

    1. You must not modify the array (assume the array is read only).
    2. You must use only constant, O(1) extra space.
    3. Your runtime complexity should be less than O(n2).
    4. 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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  • 原文地址:https://www.cnblogs.com/immjc/p/9144264.html
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