[LeetCode] First Missing Positive

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.

思路：新建一个数组B，然后用A[i]的值作为B的index，即B[A[i]] =1，然后遍历一遍B,寻找B[i]！=1的index，不过貌似我的时间复杂度是O(n)，空间也是O（n）啊。。

class Solution {public:int firstMissingPositive(int A[], int n)  {   vector<int> B;B.resize(n + 1); //cout << "n\t" << n << endl;//printArray(A, n);for(int i = 0; i < n; i++){   if(A[i] <= n && A[i] >0) {   // do A[i]'s value as B's indexB[A[i]] = 1;}   }   //printVector(B);for(size_t i = 1; i < B.size(); i++){   if(B[i] != 1)return i;}   return  n+1;}
};

为了实现O(n)的空间复杂度，只能使用数组A自身的空间，A[i]保存i+1,不停的交换，具体看注释吧，我觉得写的挺清楚的。。

class Solution {public:
#if 0int firstMissingPositive(int A[], int n) {vector<int> B;B.resize(n + 1);//cout << "n\t" << n << endl;//printArray(A, n);for(int i = 0; i < n; i++){if(A[i] <= n && A[i] >0){// do A[i]'s value as B's indexB[A[i]] = 1;}}//printVector(B);for(size_t i = 1; i < B.size(); i++){if(B[i] != 1)return i;}return  n+1;}
#endifint firstMissingPositive(int A[], int n){int i = 0;int tmp;// A[i] should store (i + 1)// it means A[0] stores 1, A[1] stores 2, ...while(i < n){// A[0~n-1] only can store 1~n;// A[i] should store i + 1// A[i] != i +1 menas A[i] should be move to others, but A[i] should be moved where?// A[i] should be move to A[i]-1// i.e.: {3, 1, 4, -1}// A[0]= 3, != (0+1), so 3 shoule be moved. moved to A[0] -1 =2, then check if A[2] and A[0] are equal.// so swap 3 and 4, then the array is {4, 1, 3 , -1}// but 4 also doesn't appear at A[0], so go on to swap, until the condition doens't meet, then i++if(A[i] <= n && A[i] >0 && A[i] != (i+1) && A[A[i] - 1] != A[i] ){tmp = A[i];A[i] = A[tmp - 1];A[tmp - 1] = tmp;}elsei++;}//printArray(A, n);for(int i = 0; i < n; i++){if(A[i] != (i+1) )return i + 1;}return n+1;}
};

转载于:https://www.cnblogs.com/diegodu/p/4285599.html

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