LintCode-61.搜索区间

搜索区间

给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。
如果目标值不在数组中，则返回[-1, -1]

样例

给出[5, 7, 7, 8, 8, 10]和目标值target=8,
返回[3, 4]

挑战

时间复杂度 O(log n)

标签

数组 排序数组 二分法

code

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) {
        // write your code here
        vector<int> result;
        int size = A.size()-1;
        if(A.empty()) {
            result.push_back(-1);
            result.push_back(-1);
        }
        else {
            result.push_back(getFirstTarget(A, target, 0, size));
            result.push_back(getLastTarget(A, target, 0, size));
        }
        return result;
    }

    int getFirstTarget(vector<int> &A, int target, int low, int high) {
        if(low > high)
            return -1;
        int mid = (low+high)/2;
        if(A[mid] == target) {
            if((mid>0 && A[mid-1]!=target) || mid==0)
                return mid;
            else
                high = mid-1;
        }
        else if(A[mid] > target)
            high = mid-1;
        else
            low = mid+1;

        return getFirstTarget(A, target, low, high);
    }

    int getLastTarget(vector<int> &A, int target, int low, int high) {
        int size = A.size()-1;
        if(low > high)
            return -1;
        int mid = (low+high)/2;
        if(A[mid] == target) {
            if((mid<size && A[mid+1]!=target) || mid==size)
                return mid;
            else
                low = mid+1;
        }
        else if(A[mid] > target)
            high = mid-1;
        else
            low = mid+1;

        return getLastTarget(A, target, low, high);
    }
};