Search for a Range
Question
- lintcode: (61) 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].
Example
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
題解
Search for a range 的題目可以拆解為找 first & last position 的題目,即要做兩次二分。由上題二分查找可找到滿足條件的左邊界,因此只需要再將右邊界找出即可。注意到在(target == nums[mid]
時賦值語句為end = mid
,將其改為start = mid
即可找到右邊界,解畢。
Java
/**
* 本代碼fork自九章算法。沒有版權歡迎轉發。
* http://www.jiuzhang.com/solutions/search-for-a-range/
*/
public class Solution {
/**
*@param A : an integer sorted array
*@param target : an integer to be inserted
*return : a list of length 2, [index1, index2]
*/
public ArrayList<Integer> searchRange(ArrayList<Integer> A, int target) {
ArrayList<Integer> result = new ArrayList<Integer>();
int start, end, mid;
result.add(-1);
result.add(-1);
if (A == null || A.size() == 0) {
return result;
}
// search for left bound
start = 0;
end = A.size() - 1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A.get(mid) == target) {
end = mid; // set end = mid to find the minimum mid
} else if (A.get(mid) > target) {
end = mid;
} else {
start = mid;
}
}
if (A.get(start) == target) {
result.set(0, start);
} else if (A.get(end) == target) {
result.set(0, end);
} else {
return result;
}
// search for right bound
start = 0;
end = A.size() - 1;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (A.get(mid) == target) {
start = mid; // set start = mid to find the maximum mid
} else if (A.get(mid) > target) {
end = mid;
} else {
start = mid;
}
}
if (A.get(end) == target) {
result.set(1, end);
} else if (A.get(start) == target) {
result.set(1, start);
} else {
return result;
}
return result;
// write your code here
}
}
源碼分析
- 首先對輸入做異常處理,數組為空或者長度為0
- 初始化
start, end, mid
三個變量,注意mid的求值方法,可以防止兩個整型值相加時溢出 - 使用迭代而不是遞歸進行二分查找
- while終止條件應為
start + 1 < end
而不是start <= end
,start == end
時可能出現死循環 - 先求左邊界,迭代終止時先判斷
A.get(start) == target
,再判斷A.get(end) == target
,因為迭代終止時target必取start或end中的一個,而end又大於start,取左邊界即為start. - 再求右邊界,迭代終止時先判斷
A.get(end) == target
,再判斷A.get(start) == target
- 兩次二分查找除了終止條件不同,中間邏輯也不同,即當
A.get(mid) == target
如果是左邊界(first postion),中間邏輯是end = mid
;若是右邊界(last position),中間邏輯是start = mid
- 兩次二分查找中間勿忘記重置
start, end
的變量值。
C++
class Solution {
/**
*@param A : an integer sorted array
*@param target : an integer to be inserted
*return : a list of length 2, [index1, index2]
*/
public:
vector<int> searchRange(vector<int> &A, int target) {
// good, fail are the result
// When found, returns good, otherwise returns fail
int N = A.size();
vector<int> fail = {-1, -1};
if(N == 0)
return fail;
vector<int> good;
// search for starting position
int lo = 0, hi = N;
while(lo < hi){
int m = lo + (hi- lo)/2;
if(A[m] < target)
lo = m + 1;
else
hi = m;
}
if(A[lo] != target)
return fail;
good.push_back(lo);
// search for ending position
lo = 0; hi = N;
while(lo < hi){
int m = lo + (hi - lo)/2;
if(target < A[m])
hi = m;
else
lo = m + 1;
}
good.push_back(lo - 1);
return good;
}
};
源碼分析
與前面題目類似,此題是將兩個子題組合起來,前半為找出"不小於target的最左元素",後半是"不大於target的最右元素",同樣的,使用開閉區間[lo, hi)仍然可以簡潔的處理各種邊界條件,僅須注意在解第二個子題"不大於target的最右元素"時,由於每次lo
更新時都至少加1,最後會落在我們要求的位置的下一個,因此記得減1回來,若直覺難以理解,可以使用一個例子在紙上推一次每個步驟就可以體會。