61. Search for a Range【medium】
61. Search for a Range【medium】
Given a sorted array of n integers, find the starting and ending position of a given target value.
If the target is not found in the array, return [-1, -1]
.
Given [5, 7, 7, 8, 8, 10]
and target value 8
,
return [3, 4]
.
O(log n) time.
解法一:
class Solution {
public:
/*
* @param A: an integer sorted array
* @param target: an integer to be inserted
* @return: a list of length 2, [index1, index2]
*/
vector<int> searchRange(vector<int> &A, int target) {
if (A.empty()) {
return vector<int>(, -);
} vector<int> result; //find first
int start = ;
int end = A.size() - ; while (start + < end) {
int mid = start + (end - start) / ;
if (A[mid] == target) {
end = mid;
}
else if (A[mid] < target) {
start = mid;
}
else if (A[mid] > target) {
end = mid;
}
} if (A[start] == target) {
result.push_back(start);
}
else if (A[end] == target) {
result.push_back(end);
}
else {
return vector<int>(, -);
} //find last
start = ;
end = A.size() - ; while (start + < end) {
int mid = start + (end - start) / ;
if (A[mid] == target) {
start = mid;
}
else if (A[mid] < target) {
start = mid;
}
else if (A[mid] > target) {
end = mid;
}
} if (A[end] == target) {
result.push_back(end);
}
else if (A[start] == target) {
result.push_back(start);
} return result;
}
};
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