leetcode26

public class Solution {
    public int RemoveDuplicates(int[] nums) {
        var len = nums.Length;
            if (len == 0)
            {
                return 0;
            }
            else
            {
                var pre = nums[0];
                var diffindex = 0;
                for (int i = 1; i < nums.Length; i++)
                {
                    if (pre != nums[i])
                    {
                        nums[++diffindex] = nums[i];
                        pre = nums[i];
                    }
                }
                return diffindex + 1;
            }
    }
}

python实现如下：

 1 class Solution:
 2     def removeDuplicates(self, nums: List[int]) -> int:
 3         s = set()
 4         n = len(nums)
 5         count = 0
 6         i,j = 0,0
 7         while i < n:
 8             if len(s) == 0 or nums[i] not in s:
 9                 s.add(nums[i])
10                 nums[j] = nums[i]
11                 i += 1
12                 j += 1
13                 count += 1
14             else:
15                 i += 1
16         return count

不实用额外的空间，实现如下：

 1 class Solution:
 2     def removeDuplicates(self, nums: 'List[int]') -> int:
 3         n = len(nums)
 4         i = 0
 5         j = i + 1
 6         count = 0
 7         while i < n:
 8             while j < n and nums[i] == nums[j]:
 9                 j += 1
10             count += 1
11             if i + 1 < n and j < n:
12                 nums[i+1] = nums[j]
13                 i += 1
14             else:
15                 return count
16         return count

python另外一种写法：

 1 class Solution:
 2     def removeDuplicates(self, nums: 'List[int]') -> int:
 3         prevalue = 0
 4         n = len(nums)
 5         count = 0
 6         i,j = 0,0
 7         while i < n:
 8             if i == 0:
 9                 prevalue = nums[i]
10                 count += 1
11                 i += 1
12                 j += 1
13             else:
14                 cur = nums[i]
15                 if cur == prevalue:#连续相同值
16                     i += 1
17                 else:
18                     nums[j] = cur
19                     count += 1
20                     prevalue = cur
21                     i += 1
22                     j += 1
23         return count

posted on 2017-04-22 10:26 Sempron2800+ 阅读(124) 评论(0) 编辑收藏举报