LeetCode--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].

思路：二分法。这道题还是二分查找法，套用经典的模版，注意两次查找，第一次更新high，找到最左边的索引，第二次更新low，找到最右边的索引。

class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
vector<int>result;
int left=-1;
int low=0,high=nums.size()-1;
while(low<=high){
int mid=(low+high)/2;
if(nums[mid]>target)
high=mid-1;
else if(nums[mid]<target)
low=mid+1;
else{
left=mid;
high=mid-1;
}
}
int right=-1;
low=0,high=nums.size()-1;
while(low<=high){
int mid=(low+high)/2;
if(nums[mid]>target)
high=mid-1;
else if(nums[mid]<target)
low=mid+1;
else{
right=mid;
low=mid+1;
}
}
result.push_back(left);
result.push_back(right);
return result;
}
};

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