First Missing Positive

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.

分析

本质上是桶排序(bucket sort)，每当 A[i]!= i+1 的时候，将A[i]与A[A[i]-1]交换，直到无法交换为止，终止条件是 A[i]== A[A[i]-1]。

代码

Java

// First Missing Positive
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
    public int firstMissingPositive(int[] nums) {
        bucket_sort(nums);

        for (int i = 0; i < nums.length; ++i)
            if (nums[i] != (i + 1))
                return i + 1;
        return nums.length + 1;
    }
    private static void bucket_sort(int[] A) {
        final int n = A.length;
        for (int i = 0; i < n; i++) {
            while (A[i] != i + 1) {
                if (A[i] < 1 || A[i] > n || A[i] == A[A[i] - 1])
                    break;
                // swap
                int tmp = A[i];
                A[i] = A[tmp - 1];
                A[tmp - 1] = tmp;
            }
        }
    }
}

C++

// First Missing Positive
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        bucket_sort(nums);

        for (int i = 0; i < nums.size(); ++i)
            if (nums[i] != (i + 1))
                return i + 1;
        return nums.size() + 1;
    }
private:
    static void bucket_sort(vector<int>& A) {
        const int n = A.size();
        for (int i = 0; i < n; i++) {
            while (A[i] != i + 1) {
                if (A[i] <= 0 || A[i] > n || A[i] == A[A[i] - 1])
                    break;
                swap(A[i], A[A[i] - 1]);
            }
        }
    }
};

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