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.
Analysis:
桶排序,每次当A[i]!= i的时候,将A[i]与A[A[i]]交换,直到无法交换位置。终止条件是 A[i]== A[A[i]]。
然后i -> 0 到n走一遍就好了。
Java
public int firstMissingPositive(int[] A) { for(int i=0;i<A.length;i++){ while(A[i]!=i+1){ if(A[i]<=0 ||A[i]>=A.length|| A[i]==A[A[i]-1]) break; int temp = A[i]; A[i] = A[A[i]-1]; A[temp-1] = temp; } } for(int i=0;i<A.length;i++){ if(A[i]!=i+1) return i+1; } return A.length+1; }
c++
int firstMissingPositive(int A[], int n) { int i=0; while(i<n){ if(A[i]!=i+1 && A[i]>=1 &&A[i]<=n && A[A[i]-1] !=A[i]) swap(A[i],A[A[i]-1]); else i++; } for(int i=0; i<n;i++){ if(A[i]!=i+1) return i+1; } return n+1; }