leetcode 442. Find All Duplicates in an Array 查找数组中的所有重复项

https://leetcode.com/problems/find-all-duplicates-in-an-array/description/

参考：http://www.cnblogs.com/grandyang/p/4843654.html

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

Example:

Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate element must exist. Assume that there is only one duplicate number, find the duplicate one.

Note:

1. You must not modify the array (assume the array is read only).
2. You must use only constant extra space.
3. Your runtime complexity should be less than O(n2).

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

这道题给了我们n+1个数，所有的数都在[1, n]区域内，首先让我们证明必定会有一个重复数，这不禁让我想起了小学华罗庚奥数中的抽屉原理(又叫鸽巢原理), 即如果有十个苹果放到九个抽屉里，如果苹果全在抽屉里，则至少有一个抽屉里有两个苹果，这里就不证明了，直接来做题吧。题目要求我们不能改变原数组，即不能给原数组排序，又不能用多余空间，那么哈希表神马的也就不用考虑了，又说时间小于O(n²)，也就不能用brute
force的方法，那我们也就只能考虑用二分搜索法了，我们在区别[1,
n]中搜索，首先求出中点mid，然后遍历整个数组，统计所有小于等于mid的数的个数，如果个数大于mid，则说明重复值在[mid+1,
n]之间，反之，重复值应在[1, mid-1]之间，然后依次类推，直到搜索完成，此时的low就是我们要求的重复值，参见代码如下：

解法一：

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int low = 1, high = nums.size() - 1;
        while (low < high) {
            int mid = low + (high - low) * 0.5;
            int cnt = 0;
            for (auto a : nums) {
                if (a <= mid) ++cnt;
            }
            if (cnt <= mid) low = mid + 1;
            else high = mid;
        }
        return low;
    }
};

经过热心网友waruzhi的留言提醒还有一种O(n)的解法，并给了参考帖子，发现真是一种不错的解法，其核心思想快慢指针在之前的题目Linked List Cycle II中就有应用，这里应用的更加巧妙一些，由于题目限定了区间[1,n]，所以可以巧妙的利用坐标和数值之间相互转换，而由于重复数字的存在，那么一定会形成环，我们用快慢指针可以找到环并确定环的起始位置，确实是太巧妙了！

解法二：

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int slow = , fast = , t = ;
        while (true) {
            slow = nums[slow];
            fast = nums[nums[fast]];
            if (slow == fast) break;
        }
        while (true) {
            slow = nums[slow];
            t = nums[t];
            if (slow == t) break;
        }
        return slow;
    }
};

类似题目：

First Missing Positive

Missing Number

Find All Numbers Disappeared in an Array

Set Mismatch

Array Nesting

参考资料：

https://leetcode.com/discuss/60830/python-solution-explanation-without-changing-input-array

https://discuss.leetcode.com/topic/29101/simple-c-code-with-o-1-space-and-o-nlogn-time-complexity

https://discuss.leetcode.com/topic/25913/my-easy-understood-solution-with-o-n-time-and-o-1-space-without-modifying-the-array-with-clear-explanation

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