First Missing Positive

问题描述

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.

解决思路

假设数组长度为n，1. 交换：如果当前元素是1-n之间的数，则将其交换到合适的位置上（如果该位置不是该元素）；

2. 扫描 + 检查。

时间复杂度为O(n)，空间复杂度为O(1).

程序

public class Solution {
    public int firstMissingPositive(int[] nums) {
		if (nums == null || nums.length == 0) {
			return 1;
		}
		int n = nums.length;
		// swap
		for (int i = 0; i < n; i++) {
			int elem = nums[i];
			if (elem < 1 || elem > n || elem == i + 1 || elem == nums[elem - 1]) {
				continue;
			}
			swap(nums, i, elem - 1);
			--i;
		}
		
		// scan and check miss
		for (int i = 0; i < n; i++) {
			if (nums[i] != i + 1) {
				return i + 1;
			}
		}
		
		return n + 1;
	}

	private void swap(int[] nums, int i, int j) {
		int tmp = nums[i];
		nums[i] = nums[j];
		nums[j] = tmp;
	}
}