[leetcode] First Missing Positive

Given an unsorted integer array, find the first missing positive integer.

For example,
Given[1,2,0]return3,
and[3,4,-1,1]return2.

Your algorithm should run in O(n) time and uses constant space.

https://oj.leetcode.com/problems/first-missing-positive/

思路：交换数组元素，使得数组中第i位存放数值(i+1)。最后遍历数组，寻找第一个不符合此要求的元素，返回其下标。整个过程需要遍历两次数组，复杂度为O(n)。

 

public class Solution {
	public int firstMissingPositive(int[] A) {
		if (A == null && A.length == 0)
			return 1;
		int n = A.length;
		int i;
		for (i = 0; i < n; i++) {
			while (A[i] > 0 && A[i] != i + 1 && A[i] <= n
					&& A[i] != A[A[i] - 1]) {
				swap(A, i, A[i] - 1);
			}

		}
		for (i = 0; i < n; i++)
			if (A[i] != i + 1)
				return i + 1;

		return n + 1;

	}

	private void swap(int[] a, int i, int j) {
		int tmp = a[i];
		a[i] = a[j];
		a[j] = tmp;

	}

	public static void main(String[] args) {
		 System.out.println(new Solution().firstMissingPositive(new int[] { 1,
		 2, 0 }));
		 System.out.println(new Solution().firstMissingPositive(new int[] { 3,
		 4, -1, 1 }));
		 System.out
		 .println(new Solution().firstMissingPositive(new int[] { 0 }));
		 System.out
		 .println(new Solution().firstMissingPositive(new int[] { 1 }));
		 System.out
		 .println(new Solution().firstMissingPositive(new int[] { 2 }));
		
		 System.out.println(new Solution()
		 .firstMissingPositive(new int[] { 0, 1 }));
		System.out.println(new Solution()
				.firstMissingPositive(new int[] { 1, 1 }));
	}

}

 

第二遍记录：

　　注意有重复元素的情况

　　注意都是合法正数情况下，返回n+1

public class Solution {
    public int firstMissingPositive(int[] A) {
        if(A==null||A.length==0)
            return 1;
        int n = A.length;
        for(int i=0;i<n;i++){
            while(A[i]>0&&A[i]<=n&&A[i]!=i+1&&A[i]!=A[A[i]-1]){
                swap(A,i,A[i]-1);
            }
        }
        for(int i=0;i<n;i++){
            if(A[i]!=i+1)
                return i+1;
        }
        
        return n+1;
        
    }
    private void swap(int[]a, int i,int j){
        int tmp= a[i];
        a[i]=a[j];
        a[j]=tmp;
    }
    
}

 

第三遍：注意交换的while循环。