Leetcode_first-missing-positive (updated c++ and python version)

地址：http://oj.leetcode.com/problems/first-missing-positive/

思路：模拟题。用一个足够长的布尔数组来存储。时间复杂度O(N), 空间复杂度都O(k).

在做题之前应该明白第一个缺失的正数应该是1~n+1之间的，如果1~n都被填满，那么应该返回n+1

如果k存在，则放在k-1的下标处，即A[k-1]，具体如下操作：

1. 如果A[k] <=0 || A[k] > n, 无效, k递增，下一次循环

2. 如果A[k] <= k+1, 直接将A[k]赋值为k+1，k递增，下一次循环

3. 否则，若A[ A[K]-1 ] 值 <=0 || >n, 说明A[ A[K]-1 ] 值无效，直接将A[ A[K]-1 ] 赋值为A[k]

4. 否则，[ A[K]-1 ]  值有效，如果A[k] 与 A[ A[k]-1 ] 不相等， 置换，(相等的话k递增继续处理，因为A[k]的值被其后的A[A[k]-1]保留了)

5. 否则，k递增下一次循环

参考代码：c++

//SECOND TRIAL, O(N) running time with constant space
class Solution {
public:
    int firstMissingPositive(int A[], int n) {
        int k = 0;
        while(k<n)
        {
            if(A[k]>0 && A[k]<=n)
            {
                if(A[k]<=k+1)
                    A[A[k]-1] = A[k++];
                else if(A[A[k]-1]<=0 || A[A[k]-1]>n)
                    A[A[k]-1] = A[k++];
                else if(A[A[k]-1]!=A[k])
                    swap(A[k], A[A[k]-1]);
                else
                    ++k;
            }
            else
                ++k;
        }
        for(int i = 0; i<n; ++i)
            if(A[i]!=i+1)
                return i+1;
        return n+1;
    }
};

python:

class Solution:
    # @param A, a list of integers
    # @return an integer
    def firstMissingPositive(self, A):
        cnt = 0
        while cnt < len(A):
            if A[cnt] > 0 and A[cnt] <= len(A):
                if A[cnt] <= cnt + 1:
                    A[ A[cnt]-1 ] = A[cnt]
                    cnt += 1
                elif A[ A[cnt]-1 ] <=0 or A[ A[cnt]-1 ]>len(A):
                    A[ A[cnt]-1 ] = A[cnt]
                    cnt += 1
                elif A[cnt] != A[ A[cnt]-1 ]:
    #better to use tmp to swap
                    #tmp = A[cnt]
                    #A[cnt] = A[A[cnt]-1]
                    #A[tmp-1] = tmp
    #Attention!! this one is OK, but A[cnt], A[A[cnt]-1] = A[A[cnt]-1], A[cnt] is wrong
                    A[A[cnt]-1], A[cnt] = A[cnt], A[A[cnt]-1]
                else:
                    cnt += 1
            else:
                cnt += 1
        
        for i in range(0, len(A)):
            if A[i] != i+1:
                return i+1
        return len(A)+1