[LeetCode] 41. First Missing Positive 首个缺失的正数

Given an unsorted integer array, find the smallest missing positive integer.

Example 1:

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

Example 2:

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

Example 3:

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

Note:

Your algorithm should run in O(n) time and uses constant extra space.

给一个没排序的整数数组，找出缺失的最小的正整数。

解法：

Java:

public class Solution {
    public int firstMissingPositive(int[] A) {
        int i = 0;
        while(i < A.length){
            if(A[i] == i+1 || A[i] <= 0 || A[i] > A.length) i++;
            else if(A[A[i]-1] != A[i]) swap(A, i, A[i]-1);
            else i++;
        }
        i = 0;
        while(i < A.length && A[i] == i+1) i++;
        return i+1;
    }
    
    private void swap(int[] A, int i, int j){
        int temp = A[i];
        A[i] = A[j];
        A[j] = temp;
    }
}

　　

Python: wo

class Solution(object):
    def firstMissingPositive(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        mx = 0
        flag = set()
        for n in nums:
            if n > mx:
                mx = n
            flag.add(n) 
        i = 1
        while i < mx:
            if i not in flag:
                return i
            i += 1
        return mx + 1  

Python:

class Solution:
    # @param A, a list of integers
    # @return an integer
    def firstMissingPositive(self, A):
        i = 0
        while i < len(A):
            if A[i] > 0 and A[i] - 1 < len(A) and A[i] != A[A[i]-1]:
                A[A[i]-1], A[i] = A[i], A[A[i]-1]
            else:
                i += 1

        for i, integer in enumerate(A):
            if integer != i + 1:
                return i + 1
        return len(A) + 1

C++:

// NOT constant space
class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int mx = 0;
        unordered_set<int> s;
        for (int num : nums) {
            if (num <= 0) continue;
            s.insert(num);
            mx = max(mx, num);
        }
        for (int i = 1; i <= mx; ++i) {
            if (!s.count(i)) return i;
        }
        return mx + 1;
    }
};

C++:

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            while (nums[i] > 0 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) {
                swap(nums[i], nums[nums[i] - 1]);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] != i + 1) return i + 1;
        }
        return n + 1;
    }
};

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Find the Duplicate Number

Find All Numbers Disappeared in an Array

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