LeetCode: First Missing Positive 解题报告

Q:

Given an unsorted integer array, find the first missing positive integer.

For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.

Your algorithm should run in O(n) time and uses constant space.

寻找数组中第一个未出现的正整数，题目本身比较常见，关键在于题目要求只能使用常数额外空间。

A:

转自Annie大神的解释

http://www.cnblogs.com/AnnieKim/archive/2013/04/21/3034631.html：

虽然不能再另外开辟非常数级的额外空间，但是可以在输入数组上就地进行swap操作。

思路：交换数组元素，使得数组中第i位存放数值(i+1)。最后遍历数组，寻找第一个不符合此要求的元素，返回其下标。整个过程需要遍历两次数组，复杂度为O(n)。

下图以题目中给出的第二个例子为例，讲解操作过程。

主页君的代码：

注意的地儿是：

1. 判断是否swap时要判目标值是不是相同（会有重复情况），否则会导致死循环。

2. 如果没找到，返回数组下标+1

 1 package Algorithms.array;
 2
 3 public class FirstMissingPositive {
 4     public static void main(String[] strs) {
 5         int[] in = {1,2,0};
 6         //int[] in = {3,4,-1,1};
 7         System.out.println(firstMissingPositive(in));
 8     }
 9
10     public static int firstMissingPositive(int[] A) {
11         if (A == null) {
12             return 0;
13         }
14
15         int len = A.length;
16         for (int i = 0; i < len; i++) {
17             // 1. The number should be in the range.
18             // 2. The number should be positive.
19             // 注意：和将要交换的值不可以相同，否则会死循环
20             while (A[i] <= len && A[i] > 0 && A[A[i] - 1] != A[i]) {
21                 swap(A, i, A[i] - 1);
22             }
23         }
24
25         for (int i = 0; i < len; i++) {
26             if (A[i] != i + 1) {
27                 return i + 1;
28             }
29         }
30
31         return len + 1;
32     }
33
34     public static void swap(int[] A, int i, int j) {
35         int tmp = A[i];
36         A[i] = A[j];
37         A[j] = tmp;
38     }
39 }

View Code

Git代码链接

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