Leetcode 34 Search for a Range

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

解题思路：采用分治的思想对数组进行折半查找，这题重点在怎么找到target在数组的左端点的位置、右端点的位置，比如判断左端点就是找对应target的下标的前一个位置所对应的值，是否小于target；同理右端点也是这样思路，然后再分别判断一下各自边界条件是啥。

 1 #include <stdio.h>
 2 
 3 #include <vector>
 4 
 5 int left_bound(std::vector<int>& nums, int target){
 6     int begin = 0;
 7     int end = nums.size() - 1;
 8     while(begin <= end){
 9         int mid = (begin + end) / 2;
10         if (target == nums[mid]){
11             if (mid == 0 || nums[mid -1] < target){
12                 return mid;
13             }
14             end = mid - 1;
15         }
16         else if (target < nums[mid]){
17             end = mid - 1;
18         }
19         else if (target > nums[mid]){
20             begin = mid + 1;
21         }
22     }
23     return -1;
24 }
25 
26 int right_bound(std::vector<int>& nums, int target){
27     int begin = 0;
28     int end = nums.size() - 1;
29     while(begin <= end){
30         int mid = (begin + end) / 2;
31         if (target == nums[mid]){
32             if (mid == nums.size() - 1 || nums[mid + 1] > target){
33                 return mid;
34             }
35             begin = mid + 1;
36         }
37         else if (target < nums[mid]){
38             end = mid - 1;
39         }
40         else if (target > nums[mid]){
41             begin = mid + 1;
42         }
43     }
44     return -1;
45 }
46 
47 class Solution {
48 public:
49     std::vector<int> searchRange(std::vector<int>& nums, int target) {
50         std::vector<int> result;
51         result.push_back(left_bound(nums, target));
52         result.push_back(right_bound(nums, target));
53         return result;
54     }
55 };
56 
57 int main(){
58     int test[] = {5, 7, 7, 8, 8, 8, 8, 10};
59     std::vector<int> nums;
60     Solution solve;
61     for (int i = 0; i < 8; i++){
62         nums.push_back(test[i]);
63     }
64     for (int i = 0; i < 12; i++){
65         std::vector<int> result = solve.searchRange(nums, i);
66         printf("%d : [%d, %d]\n",i , result[0], result[1]);
67     }
68     return 0;
69 }

通过~

 

posted @ 2018-03-24 21:28  Hwangzhiyoung  阅读(140)  评论(0)  编辑  收藏  举报