First Missing Positive
描述
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.
分析
本质上是桶排序(bucket sort)，每当 A[i]!= i+1 的时候，将A[i]与A[A[i]-1]交换，直到无法交换为止，终止条件是 A[i]== A[A[i]-1]。
代码
// First Missing Positive
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
int firstMissingPositive(vector<int>& nums) {
bucket_sort(nums);
for (int i = 0; i < nums.size(); ++i)
if (nums[i] != (i + 1))
return i + 1;
return nums.size() + 1;
}
private:
static void bucket_sort(vector<int>& A) {
const int n = A.size();
for (int i = 0; i < n; i++) {
while (A[i] != i + 1) {
if (A[i] <= 0 || A[i] > n || A[i] == A[A[i] - 1])
break;
swap(A[i], A[A[i] - 1]);
}
}
}
};