34. Search for a Range (Array; Divide-and-Conquer)

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：先二分法找最左端，再二分法找最右端。保证稳定排序。具体实现：

相同元素返回最左元素：start总是在<target位置，所以从-1开始

相同元素返回最右元素：end总是在>target位置，所以从n开始

二者的结束条件都是start+1==end

class Solution {
public:
vector< int > searchRange(int A[], int n, int target) {
result.clear();
binarySearchLeft(A,-1,n,target);
if(result[0]!=-1) binarySearchRight(A,-1,n,target);
return result;
}

void binarySearchLeft(int A[], int start, int end, int target){ //find the first value == target
if (start+1 == end){
if(end >= n || A[end]!=target){
result.push_back(-1);
result.push_back(-1);
}
else result.push_back(end);
return;
}

int mid = start + ((end-start)>>1);

if (A[mid] < target)
binarySearchLeft(A, mid, end, target);
else
binarySearchLeft(A, start, mid, target);
}

void binarySearchRight(int A[], int start, int end, int target){ //find the first value == target
if (start+1 == end){
result.push_back(start);
return;
}

int mid = start + ((end-start)>>1);

if (A[mid] > target)
binarySearchRight(A, start, mid, target);
else
binarySearchRight(A, mid, end, target);
}
private:
vector< int >  result;
};