标签：correct app pac missing counter const stat make use
一、题目
1、描述
2、题意
返回无序数组中缺失的第一个正整数。
二、解答
1、思路：
① 数组进行排序，下标从数值为正整数开始；
② 若后面数值与前面数值重复，下标+1；
③ 返回 第一个缺失的数值，跳出循环
// 41. First Missing Positive public static int firstMissingPositive(int[] nums) { int len = nums.length; if(len == 0) return 1; Arrays.sort(nums); int index = 0; while(index < len && nums[index] <= 0) index++; for(int i = 1; i <= len + 1; i++) { if(index < len) { if(nums[index++] != i) return i; else { while(index < len && nums[index-1] == nums[index]) index++; } } else { return i; } } return 0; }
方法二：
The key here is to use swapping to keep constant space and also make use of the length of the array, which means there can be at most n positive integers. So each time we encounter an valid integer, find its correct position and swap. Otherwise we continue.
public class Solution { public int firstMissingPositive(int[] A) { int i = 0; while(i < A.length){ if(A[i] == i+1 || A[i] <= 0 || A[i] > A.length) i++; else if(A[A[i]-1] != A[i]) swap(A, i, A[i]-1); else i++; } i = 0; while(i < A.length && A[i] == i+1) i++; return i+1; } private void swap(int[] A, int i, int j){ int temp = A[i]; A[i] = A[j]; A[j] = temp; } }
三、总结
方法二还需探究。 A[A[i]-1] != A[i] 仅仅是避免死循环？
标签：correct app pac missing counter const stat make use
原文地址：http://www.cnblogs.com/skillking/p/7093183.html
