第九周演算法分析與設計: Search for a Range
阿新 • • 發佈:2019-02-11
問題描述:
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 ofO(logn) .
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]
.
問題來自此處
解答思路:
一看到題目要求時間複雜度是二分法
啦!
我的思路是先找到目標數,然後往目標數的左右分別延伸查詢,定下相同數所在的上下界。
vector<int> searchRange(vector<int>& nums, int target) {
vector<int> result(2,-1); //初始化為[-1,-1]
int size = nums.size();
if (size == 0){
return result;
}
int low = 0, high = size - 1, med;
while(low <= high){
med = (low + high) / 2;
if(nums[med] == target){
int left = find_bound(nums,0,med,target);
int right = find_bound(nums,size,med,target);
result.clear();
result.push_back(left);
result.push_back(right);
return result;
}
else if(nums[med] < target){
low = med + 1;
}
else{
high = med - 1;
}
}
return result;
}
int find_bound(vector<int> nums,int numb,int med,int target){
int res = med;
if(numb<=med){
for(int i = med;i >= numb;i--){
if(nums[i] == target)
res = i;
if(nums[i] < target)
break;
}
}
else{
for(int i = med;i < numb;i++){
if(nums[i] == target)
res = i;
if(nums[i] > target)
break;
}
}
return res;
}
寫得很煞筆。。我以為執行時間的排名鐵定會被排在最後那批。。結果居然在中間……有毒有毒~