LeetCode 之 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]

.

有序数组中的查找，使用二分查找，由于是要查重复元素的上下边界，所以要不止一次的二分查找，根据网上一篇博文的思路是用两次二分查找分别找到左右边界，需要注意到：在寻找左边界时，low要一直寻找target元素，在寻找右边界时，low遇到target要向右移动，直到不等于target。代码如下：

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> ans={-1,-1};
int length=nums.size();
if(length==0) return ans;
int llow=0,lhigh=length-1;
while(llow<=lhigh){
int mid=(llow+lhigh)/2;
if(nums[mid]<target)
llow=mid+1;
else
lhigh=mid-1;
}
int rlow=0,rhigh=length-1;
while(rlow<=rhigh){
int mid=(rlow+rhigh)/2;
if(nums[mid]<=target)
rlow=mid+1;
else
rhigh=mid-1;
}
if(llow<=rhigh){
ans[0]=llow;
ans[1]=rhigh;
}
return ans;
}
};