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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.
Note the problem is to find first missing integer, ie, if input is [4,5,7,8], we should return 1.
Naive way is to traverse from 1 to length, find whether they are in original input set. Time complexity O(n), space cost O(n).
1 public class Solution { 2 public int firstMissingPositive(int[] nums) { 3 int length = nums.length; 4 Set<Integer> set = new HashSet<Integer>(); 5 for (int i = 0; i < length; i++) 6 set.add(nums[i]); 7 int result = 1; 8 for (int i = 1; i <= length; i++) { 9 if (!set.contains(i)) { 10 result = i; 11 break; 12 } else { 13 // Should increase here 14 // Consider input [1] 15 result++; 16 } 17 } 18 return result; 19 } 20 }
Key is to put i on position (i - 1). Time complexity O(n), space cost O(1).
1 public class Solution { 2 public int firstMissingPositive(int[] nums) { 3 int length = nums.length; 4 // Put i on (i - 1) position 5 // Consider three conditions: 1. nums[i] <= 0 2. nums[i] > length 3. duplicates 6 for (int i = 0; i < length; i++) { 7 int target = nums[i]; 8 if (target != i + 1) { 9 // nums[i] <= 0 nums[i] > length 10 if (target <= 0 || target > length) 11 continue; 12 // Duplicate 13 if (nums[target - 1] == target) { 14 nums[i] = 0; 15 continue; 16 } 17 // Swap target with nums[target - 1] 18 int tmp = nums[target - 1]; 19 nums[target - 1] = target; 20 nums[i] = tmp; 21 // Still need to check nums[i] after swap; 22 i--; 23 } 24 } 25 26 int result = length + 1; 27 // Check 28 for (int i = 0; i < length; i++) { 29 if (nums[i] != i + 1) { 30 result = i + 1; 31 break; 32 } 33 } 34 return result; 35 } 36 }
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原文地址：http://www.cnblogs.com/ireneyanglan/p/4870597.html
