sorting/bucket-sort/first-missing-positive
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2023-12-01
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)
public class Solution {
public int firstMissingPositive(int[] nums) {
bucket_sort(nums);
for (int i = 0; i < nums.length; ++i)
if (nums[i] != (i + 1))
return i + 1;
return nums.length + 1;
}
private static void bucket_sort(int[] A) {
final int n = A.length;
for (int i = 0; i < n; i++) {
while (A[i] != i + 1) {
if (A[i] < 1 || A[i] > n || A[i] == A[A[i] - 1])
break;
// swap
int tmp = A[i];
A[i] = A[tmp - 1];
A[tmp - 1] = tmp;
}
}
}
}
// 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]);
}
}
}
};