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448-找到所有数组中消失的数字

LeetCode

发布日期: 2022-02-22

LeetCode 448. Find All Numbers Disappeared in an Array

Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]

Example 2:

Input: nums = [1,1]
Output: [2]

method

用哈希表存储所有数，然后从1到n开始查找看哪个数不存在

vector<int> findDisappearedNumbers(vector<int>& nums) {
    vector<int> res;
    unordered_set<int> st;
    for (auto n : nums) st.insert(n);
    for (int i = 1; i <= nums.size(); i++) {
        if (!st.count(i)) res.push_back(i);
    }
    return res;
}

空间复杂度：$O(n)$

改进：直接以原数组作为哈希表，如果某个数x存在，就让nums[x-1]+n，这样加一遍之后，所有元素的范围会由[1,n]变成[n+1,2n]，所以小于这个范围的数的index+1就是缺失的数

可能x已经被加过了，所以需要取模，nums[(x-1)%len]+n

vector<int> findDisappearedNumbers(vector<int>& nums) {
    int len = nums.size();
    for (auto n : nums) {
        int index = (n - 1) % len;
        nums[index] += len;
    }
    vector<int> res;
    for (int i = 0; i < nums.size(); i++) {
        if (nums[i] <= len) res.push_back(i + 1);
    }
    return res;
}