leetcode 034 —— Search for a Range

Given a sorted array of integers, 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]

.

思路：二分法先查找到任意target所在的位置，再往两边搜索边界

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> res;
int pos = posFind(nums, target, 0, nums.size() - 1);
if (pos == -1){
res.push_back(-1);
res.push_back(-1);
return res;
}
int l = pos, r = pos;
while (nums[l] == target&&l>=0)
l--;
while (nums[r] == target&&r <= nums.size() - 1)
r++;

res.push_back(l+1);
res.push_back(r - 1);

return res;
}
int posFind(vector<int> &nums, int target, int l, int r){
int mid;
while (l <= r){
mid = (l + r) / 2;
if (nums[mid] == target)
return mid;
else if (nums[mid] < target)
l = mid + 1;
else
r = mid - 1;
}
return -1;
}
};