最后一遍-L41-First Missing Positive
描述：
Given an unsorted integer array nums, return the smallest missing positive integer.
You must implement an algorithm that runs in O(n) time and uses constant extra space.
Example 1:
Input: nums = [1,2,0]
Output: 3
Explanation: The numbers in the range [1,2] are all in the array.
Example 2:
Input: nums = [3,4,-1,1]
Output: 2
Explanation: 1 is in the array but 2 is missing.
Example 3:
Input: nums = [7,8,9,11,12]
Output: 1
Explanation: The smallest positive integer 1 is missing.
Constraints:
1 <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
快慢指针的变化：Floyd Cycle
public class L41 {
public static void main(String[] args) {
System.out.println("Keep Moving !! ");
int value = firstMissingPositive(new int[]{1,2,0});
System.out.println(value);
value = firstMissingPositive(new int[]{3,4,-1,1});
System.out.println(value);
value = firstMissingPositive(new int[]{7,8,9,11,12});
System.out.println(value);
value = firstMissingPositive(new int[]{2,1});
System.out.println(value);
}
/**
* 发现最小的正数,关键在于想到这中方法
* */
public static int firstMissingPositive(int[] a) {
int i=0;
while(i < a.length){
if(a[i]< a.length && a[i] >0 && a[i] != i+1){
swap(a,a[i]-1,i);
continue;
}
i++;
}
i=0;
while(i < a.length && a[i] == i+1) i++;
return i+1;
}
private static void swap(int[] a, int i, int j) {
int tmp = a[j];
a[j] = a[i];
a[i] = tmp;
}
}