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.
采用hash的思路，将A[i]的当做索引，标记对应的A[A[i]-1]为-， 最后遍历，为正的则为缺失的
int firstMissingPositive(int A[], int n) { for (int i=0; i<n; i++) { if (A[i] <= 0) A[i] = n+2; } for (int i=0; i<n; i++) { int nTmp = abs(A[i]); if (nTmp<1 || nTmp>n) continue; if (A[nTmp-1] > 0) A[nTmp-1] *= -1; } for (int i=0; i<n; i++) { if (A[i] > 0) return i+1; } return n+1; }
原文：http://www.cnblogs.com/zhhwgis/p/3958377.html