LeetCode: Search for a Range [033]

【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].

【题意】

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)

【思路】

使用二分查找，分别确定第一个和最后一个的位置。

【代码】

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;
    }
};

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