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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 index B[A[i]] = 1; } } //printVector(B); for(size_t i = 1; i < B.size(); i++) { if(B[i] != 1) return i; } return n+1; } };
[LeetCode] First Missing Positive
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原文地址：http://www.cnblogs.com/diegodu/p/4285599.html
