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。