class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target)
    {
        vector<int> r(2, -1);
        if (!nums.size())
        {
            return r;//为空时,返回[-1,-1]        
        }
        int l = 0, h = nums.size() - 1, m = 0;
        if (target < nums[l] || nums[h] < target)
        {
            return r; //小于最小或大于最大返回[-1,-1]        
        }
        while (l <= h)
        {
            m = (l + h) / 2;
            if (nums[m] == target)
            {
                //命中m时                
                int i = m;
                while (nums[i] == target&&i < nums.size())
                {
                    i++;//寻找nums中与target相等值的右界                
                }
                r[1] = i - 1;
                while (nums[m] == target&&m >= 0)
                {
                    m--;//寻找nums中与target相等值的左界                
                }
                r[0] = m + 1;
                return r;
            }
            else if (nums[m] < target)
            {
                l = m + 1;
            }
            else
            {
                h = m - 1;
            }
        }
        return r;//没找到target,返回[-1,-1]    
    }
};

 

补充一个python的实现:

 1 class Solution:
 2     def searchRange(self, nums: 'List[int]', target: 'int') -> 'List[int]':
 3         n = len(nums)
 4         i = 0
 5         j = n - 1
 6         if i == j:
 7             if nums[i] == target:
 8                 return [0,0]
 9             else:
10                 return [-1,-1]
11         begin = 0
12         end = n - 1
13         while i < j:
14             if nums[i] == target:
15                 begin = i
16                 end = i
17                 while end < n and nums[end] == target:
18                     end += 1
19                 return [begin,end-1]
20             if nums[j] == target:
21                 end = j
22                 begin = j
23                 while begin >= 0 and nums[begin] == target:
24                     begin -= 1
25                 return [begin+1,end]
26             mid = i + (j - i) // 2
27             if nums[mid] == target:
28                 begin = mid
29                 while begin >= 0 and nums[begin] == target:
30                     begin -= 1
31                 end = mid
32                 while end < n and nums[end] == target:
33                     end += 1
34                 return [begin+1,end-1]
35             elif nums[mid] < target:
36                 i = mid + 1
37             else:
38                 j = mid - 1
39         return [-1,-1]
40                 
41         

 

posted on 2018-10-06 16:55  Sempron2800+  阅读(127)  评论(0编辑  收藏  举报