LeetCode-First Missing Positive
2014-09-06 11:44
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题目:https://oj.leetcode.com/problems/first-missing-positive/
Given an unsorted integer array, find the first missing positive integer.
For example,
Given
and
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] (下标值要减1)交换,直到无法交换为止,终止条件是A[i]== A[A[i]-1]。作用就是把A[i]放到小标为A[i]-1的位置,遇到负值或大于n的值则退出,最后扫描数组,若A[i]!=i+1,说明在A[i]处不连续。
源码:Java版本
算法分析:时间复杂度O(n),空间复杂度O(1)
注:欢迎各位补充时间复杂度的分析过程。
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] (下标值要减1)交换,直到无法交换为止,终止条件是A[i]== A[A[i]-1]。作用就是把A[i]放到小标为A[i]-1的位置,遇到负值或大于n的值则退出,最后扫描数组,若A[i]!=i+1,说明在A[i]处不连续。
源码:Java版本
算法分析:时间复杂度O(n),空间复杂度O(1)
public class Solution { public int firstMissingPositive(int[] A) { bucketSort(A); for(int i=0;i<A.length;i++) { if(A[i]!=i+1) { return i+1; } } return A.length+1; } private void bucketSort(int[] A) { for(int i=0;i<A.length;i++) { while(A[i]!=i+1) { if(A[i]<=0 || A[i]>A.length || A[i]==A[A[i]-1]) { break; } swap(A,i,A[i]-1); } } } private void swap(int[] A,int x,int y) { int temp=A[x]; A[x]=A[y]; A[y]=temp; } }
注:欢迎各位补充时间复杂度的分析过程。
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