【一天一道LeetCode】#34. Search for a Range

一天一道LeetCode系列

（一）题目

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].

（二）解题

/*
首先二分法查找目标值
然后沿着目标值左右延伸，依次找到相同值得左右边界
*/
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int i = 0;
int j = nums.size()-1;
int idx = -1;
vector<int> ret;
while(i <= j)
{
int mid = (i+j)/2;
if(nums[mid] == target)
{
idx = mid;
break;
}
else if(nums[mid]>target)
{
j = mid-1;
}
else if(nums[mid]<target)
{
i = mid+1;
}
}
if(idx!=-1){
int k = idx;
while(k>=0 && nums[k] == target) k--;
int m = idx;
while(m<nums.size()&&nums[m] == target) m++;
ret.push_back(k+1);
ret.push_back(m-1);
}
else{
ret.push_back(-1);
ret.push_back(-1);
}
return ret;
}
};