[3]442. Find All Duplicates in an Array/[3]448. Find All Numbers Disappeared in an Array(Java)-优快云博客

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Find All Duplicates in an Array

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]

注意：输入数组中的数值的取值范围为1到n。

思路：引入索引index与nums[i]的映射函数，当第一次遇见index位置的nums[index]，将nums[index]反转置为负数，当出现与前面出现过的nums[i]值相同的nums[i]，则由映射函数算出相同的index，对应到相同的nums[index]，若此时nums[index]为一个负数，则nums[i]曾经出现过，将nums[i]加入到res数组中。

class Solution {
    public List<Integer> findDuplicates(int[] nums) {
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < nums.length; i ++) {
            int index = Math.abs(nums[i]) - 1; // nums[i]的取值范围为1到n, nums[i] - 1的取值范围为0到n-1，可作为下标索引
            if (nums[index] > 0) {
                nums[index] = - nums[index];
            } else res.add(index + 1);
        }
        return res;
    }
}

Find All Numbers Disappeared in an Array

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

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

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

Output:
[5,6]

class Solution {
    public List<Integer> findDisappearedNumbers(int[] nums) {
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < nums.length; i ++) {
            int index = Math.abs(nums[i]) - 1;
            if (nums[index] > 0) 
                nums[index] = -nums[index];    
        }
        
        for (int i = 0; i < nums.length; i ++) 
            if (nums[i] > 0) res.add(i + 1); // 相当于上面函数的nums[index] > 0，求i
        return res;
    }
}