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41.first-missing-positive

Statement

Metadata

给你一个未排序的整数数组 nums ,请你找出其中没有出现的最小的正整数。

请你实现时间复杂度为 O(n) 并且只使用常数级别额外空间的解决方案。

 

示例 1:

输入:nums = [1,2,0]
输出:3

示例 2:

输入:nums = [3,4,-1,1]
输出:2

示例 3:

输入:nums = [7,8,9,11,12]
输出:1

 

提示:

  • 1 <= nums.length <= 5 * 105
  • -231 <= nums[i] <= 231 - 1

Metadata

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.

 

Example 1:

Input: nums = [1,2,0]
Output: 3

Example 2:

Input: nums = [3,4,-1,1]
Output: 2

Example 3:

Input: nums = [7,8,9,11,12]
Output: 1

 

Constraints:

  • 1 <= nums.length <= 5 * 105
  • -231 <= nums[i] <= 231 - 1

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class Solution {
public:
    int firstMissingPositive(vector<int> &nums) {
        int n = nums.size();
        nums.push_back(INT_MAX);

        for (auto &c : nums) {
            if (c <= 0) {
                c = INT_MAX;
            }
        }

        for (auto &c : nums) {
            if (abs(c) <= n && nums[abs(c)] > 0) {
                nums[abs(c)] *= -1;
            }
        }

        for (int i = 1; i <= n; i++) {
            if (nums[i] > 0) {
                return i;
            }
        }

        return n + 1;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: January 15, 2023
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