(leetcode)Search for a Range

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         vector<int> ans(2);
 5         if(nums.size() == 0) return ans;
 6         int i = 0;
 7         int j = nums.size()-1;
 8         while(i <=j )
 9         {
10             while(i <= j && nums[i] != target)
11             {
12                 ++i;
13             }
14             while(i <= j && nums[j] != target)
15             {
16                 --j;
17             }
18             if(i<=j)
19             {
20                 ans[0] = i;
21                 ans[1] = j;
22                 return ans;
23             }
24         }
25         ans[0] = -1;
26         ans[1] = -1;
27         return ans;
28     }
29 };

 1 class Solution {
 2 public:
 3     vector<int> searchRange(vector<int>& nums, int target) {
 4         vector<int> ans(2,-1);
 5         if (nums.size() == 0) {
 6               return ans;
 7         }
 8           
 9         int len = nums.size();
10         int left = 0; 
11         int right = len - 1;
12          
13          // so when loop end, there will be 2 elements in the array.
14          // search the left bound.
15         while (left < right - 1) {
16             int mid = left + (right - left) / 2;
17             if (target == nums[mid]) {
18              // 如果相等，继续往左寻找边界
19              right = mid;
20             } else if (target > nums[mid]) {
21              // move right;
22              left = mid;
23             } else {
24              right = mid;
25             }
26         }
27          
28         if (nums[left] == target) {
29          ans[0] = left;
30         } else if (nums[right] == target) {
31          ans[0] = right;
32         } else {
33          return ans;
34         }
35          
36         left = 0; 
37         right = len - 1;
38         // so when loop end, there will be 2 elements in the array.
39         // search the right bound.
40         while (left < right - 1) {
41             int mid = left + (right - left) / 2;
42             if (target == nums[mid]) {
43             // 如果相等，继续往右寻找右边界
44              left = mid;
45             } else if (target > nums[mid]) {
46             // move right;
47              left = mid;
48             } else {
49              right = mid;
50             }
51         }
52          
53         if (nums[right] == target) {
54             ans[1] = right;
55         } else if (nums[left] == target) {
56             ans[1] = left;
57         } else {
58             return ans;
59         }
60         
61         return ans;
62     }
63 };
64