[LintCode]61.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]

思路：二分查找

class Solution {public:    /*     * @param A: an integer sorted array     * @param target: an integer to be inserted     * @return: a list of length 2, [index1, index2]     */    vector<int> searchRange(vector<int> A, int target) {        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;    }    //找左边界   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);    }    //找右边界    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);    }};