编写一段python代码，解决以下问题：Given an unsorted integer array nums, return the smallest missing positive integer. You must implement an algorithm that runs in O(n) time and uses constant extra space.

这是给出的解决方案：

def findSmallestMissingPositive(nums):
    # 将所有小于等于 0 的数替换为 len(nums) + 1
    for i in range(len(nums)):
        if nums[i] <= 0:
            nums[i] = len(nums) + 1

    # 将所有在范围内的数替换为负数
    for i in range(len(nums)):
        num = abs(nums[i])
        if num <= len(nums):
            nums[num - 1] = -abs(nums[num - 1])

    # 找到第一个正数，它就是缺失的最小正数
    for i in range(len(nums)):
        if nums[i] > 0:
            return i + 1

    # 如果数组中的所有数都是负数，则返回 len(nums) + 1
    return len(nums) + 1

这个算法在时间复杂度上做到了 O(n)，因为它只遍历了数组 nums 一次。它还使用了常数的额外空间，因为它只使用了一些变量来存储数据。

相关问题

写出这个题的代码：You are given an unsorted permutation p1,p2,…,pn. To sort the permutation, you choose a constant k (k≥1) and do some operations on the permutation. In one operation, you can choose two integers i, j (1≤j<i≤n) such that i−j=k, then swap pi and pj. What is the maximum value of k that you can choose to sort the given permutation? A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array). An unsorted permutation p is a permutation such that there is at least one position i that satisfies pi≠i. Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104). The description of the test cases follows. The first line of each test case contains a single integer n (2≤n≤105) — the length of the permutation p. The second line of each test case contains n distinct integers p1,p2,…,pn (1≤pi≤n) — the permutation p. It is guaranteed that the given numbers form a permutation of length n and the given permutation is unsorted. It is guaranteed that the sum of n over all test cases does not exceed 2⋅105. Output For each test case, output the maximum value of k that you can choose to sort the given permutation. We can show that an answer always exists.

#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        int n;
        scanf("%d", &n);
        vector<int> p(n);
        for(int i = 0; i < n; i++) {
            scanf("%d", &p[i]);
        }
        int ans = 0;
        for(int i = 0; i < n; i++) {
            if(abs(p[i] - i) > 1) { // 找到一个需要交换的位置
                ans = max(ans, abs(p[i] - i)); // 更新最大的 k
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}

写一段c++代码You are given an unsorted permutation p_1, p_2, \ldots, p_np 1 ​ ,p 2 ​ ,…,p n ​ . To sort the permutation, you choose a constant kk (k \ge 1k≥1) and do some operations on the permutation. In one operation, you can choose two integers ii, jj (1 \le j < i \le n1≤j<i≤n) such that i - j = ki−j=k, then swap p_ip i ​ and p_jp j ​ . What is the maximum value of kk that you can choose to sort the given permutation? A permutation is an array consisting of nn distinct integers from 11 to nn in arbitrary order. For example, [2, 3, 1, 5, 4][2,3,1,5,4] is a permutation, but [1, 2, 2][1,2,2] is not a permutation (22 appears twice in the array) and [1, 3, 4][1,3,4] is also not a permutation (n = 3n=3 but there is 44 in the array). An unsorted permutation pp is a permutation such that there is at least one position ii that satisfies p_i \ne ip i ​  =i.

Here's the C++ code to solve the problem:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n;
    cin >> n;
    vector<int> p(n);
    for (int i = 0; i < n; i++) {
        cin >> p[i];
    }
    int max_k = n / 2;  // maximum value of k
    for (int k = max_k; k >= 1; k--) {
        for (int i = 0; i < k; i++) {  // sort each k-length block separately
            for (int j = i+k; j < n; j += k) {
                if (p[j] < p[j-k]) {
                    swap(p[j], p[j-k]);
                }
            }
        }
        if (is_sorted(p.begin(), p.end())) {  // check if the permutation is sorted
            cout << k << endl;
            break;
        }
    }
    return 0;
}

The code reads in the input permutation, then iterates over the possible values of k in descending order from n/2 to 1. For each value of k, it sorts each k-length block separately using a simple bubble sort, and checks if the entire permutation is sorted. If it is, then it outputs the value of k and exits. If no value of k sorts the permutation, then it outputs -1.