Leetcode: Search for a Range
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].
思路:
1. 又是二分法
2. 二分法的框架, 需要考虑的位置有 3 个, 在代码中我标了出来, 分别为 q1, q2, q3
3. q1 是取 <= 还是取 <. 我的经验是, 若是题目要求找到 target, 那么就用 <=, 否则用 <. 我记得在二分搜索题时, 都是用 < 的
4. q2 比较容易, 考虑当 low == high 时, 我们希望游标往哪里走
5. q3, 返回 low/high. q3 的选取与 q2 有关. 还是需要考虑当 low == high 时, 游标会往哪走
代码
class Solution {
public:
vector<int> searchRange(int A[], int n, int target) {
vector<int> res;
if(n <= 0)
return res;
int leftIndex = lSearch(A, n, target);
int rightIndex = rSearch(A, n, target);
res.push_back(leftIndex);
res.push_back(rightIndex);
return res;
}
int lSearch(int A[], int n, int target) {
int low = 0, high = n-1;
while(low <= high) { // q1
int mid = (low+high)>>1;
if(A[mid] < target) { // q2
low = mid+1;
}else{
high = mid-1;
}
}
if(A[low] != target)
return -1;
return low; // q3
}
int rSearch(int A[], int n, int target) {
int low = 0, high = n-1;
while(low <= high) { // q1
int mid = (low+high)>>1;
if(A[mid] > target) { // q2
high = mid-1;
}else{
low = mid +1;
}
}
if(A[high] != target)
return -1;
return high; // q3
}
};