Algorithm-Arrays-5 Repeat and Missing Number Array

1. 问题

You are given a read only array of n integers from 1 to n.

Each integer appears exactly once except A which appears twice and B
which is missing.

Return A and B.

Note: Your algorithm should have a linear runtime complexity. Could
you implement it without using extra memory?

Note that in your output A should precede B.

Example:

Input:[3 1 2 5 3]

Output:[3, 4]

A = 3, B = 4

给定一个数组，数组大小为n, 且该数组中存储的数据为1到n. 除了有一个重复的数字，和一个遗失的之外，每个其他数字仅仅出现一次。编写程序实现该算法。

2. 思路

可以用2个数组，大小为数组的长度，分别存储的是重复值的位置标志和遗失的值的位置索引。

public class Solution {    // DO NOT MODIFY THE LIST    public ArrayList<Integer> repeatedNumber(final List<Integer> a) {        boolean repeat[] = new boolean[a.size()];        boolean missing[] = new boolean[a.size()];        ArrayList<Integer> result = new ArrayList<>();        int repeatNum = 0;        for (int val : a) {            if (repeat[val - 1] == false)                repeat[val - 1] = true;            else                repeatNum = val;        }        int missingNum = 0;        for (int val : a) {            missing[val - 1] = true;        }        for (int i = 0; i < a.size(); i++) {            if (!missing[i])                missingNum = i + 1;        }        result.add(repeatNum);        result.add(missingNum);        return result;    }}

3. 参考答案–另一种解法

参考答案解法：简单粗暴。首先将数组排序；然后找到重复的数字，只需便利一层循环即可得到；遗失的数字仅需通过等差数列求和公式，简单相减即可得到。

public class Solution {    // DO NOT MODIFY THE LIST    public ArrayList<Integer> repeatedNumber(final List<Integer> A) {        ArrayList<Integer> res = new ArrayList<>();        Collections.sort(A);        int n = A.size();        int rep = -1;        int miss = -1;        long sum = A.get(0);        for (int i = 1; i < n; i++) {            if (A.get(i).intValue() == A.get(i - 1).intValue()) {                rep = A.get(i);            }            sum += A.get(i);        }        miss = (int) ((1L * n * (1L * n + 1)) / 2 - sum + rep);        res.add(rep);        res.add(miss);        return res;    }}