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].
给定一个有序数组，其中包含重复数值。给定一个目标target, 返回第一个target值和最后一个target值出现的索引位置。时间复杂度保证O(logn)
class Solution {
public:
//获得第一个值的位置
int getFirst(int A[], int n, int target){
int low=0, high=n-1;
while(low<=high){
int mid=(low+high)/2;
if(A[mid]==target){
if(mid==0||A[mid-1]!=A[mid])return mid;
else high=mid-1;
}
else if(target<A[mid])high=mid-1;
else low=mid+1;
}
return -1;
}
//获得最后一个值的位置
int getLast(int A[], int n, int target){
int low=0, high=n-1;
while(low<=high){
int mid=(low+high)/2;
if(A[mid]==target){
if(mid==n-1||A[mid+1]!=A[mid])return mid;
else low=mid+1;
}
else if(target<A[mid])high=mid-1;
else low=mid+1;
}
return -1;
}
vector<int> searchRange(int A[], int n, int target) {
vector<int> result(2, -1);
if(n==0)return result;
result[0]=getFirst(A, n, target);
result[1]=getLast(A, n, target);
return result;
}
};LeetCode: Search for a Range [033],布布扣,bubuko.com
LeetCode: Search for a Range [033]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/26229829
