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.
解題思路:數組總共有n個數,若都是連續的,最大的數是n,也就是說返回的值最大為n+1,遍歷數組,如果當前的數num[i],
如果num[i]<=0或者num[i]超過了n,則第一個丟掉的整數必然在num[i]的前面,將num[i]和num[n-1]互換,n--;
如果1<=num[i]<=n,表明這個數可能是好的,將他放在正確的位置swap(num[i],num[num[i]-1])
如果num[i]==i+1,則表明該數字在正確的位置,i++
class Solution { public:int firstMissingPositive(int A[], int n) {int m = n;for(int i=0;i<m;i++){if(A[i] == i+1)continue;else{if(A[i] > m || A[i] == 0 || A[i] < 0 || A[i] <= i){if(i==m-1)return m;swap(A[i],A[m-1]);m--;i--;}else{if(A[i] == A[A[i]-1]){swap(A[i],A[m-1]);m--;i--;}else{swap(A[i],A[A[i]-1]);i--;}}}}return m+1;} };
?