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]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

// 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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