LeetCode: Search for a Range

思路：

遍历数组，碰到第一个元素等于目标值时，就是起始位置，每次更新终点位置（数组元素等于目标值的终点位置）

不过这是一种trivial的解法（虽然能够被accept），题目要求时间复杂度是O(log n)，所以必须是二分法。

二分法分两步，首先找到最左边的目标值，然后找到最右边的坐标值，最后汇总一下就可以了。

trivial code:

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ret;
        int lastIndex = 0;
        bool flag = false;
        for(int i = 0;i < n;i++)
            if(A[i] == target){
                if(!flag) ret.push_back(i);
                lastIndex = i;
                flag = true;
            }
        if(!flag){
            ret.push_back(-1);
            ret.push_back(-1);
        }
        else
            ret.push_back(lastIndex);
        return ret;
    }
};

二分法code:

class Solution {
public:
    int binaryLeftSearch(int A[],int left,int right,int target){
        int originalLeft = left, originalRight = right;
        while(left <= right){
            int mid = (left + right)/2;
            if(A[mid] == target){
                if(mid == originalLeft || A[mid-1] != target)
                    return mid;
                else
                    right = mid-1;
            }
            else if(A[mid] > target)
                right = mid - 1;
            else
                left = mid + 1;
        }
        return -1;
    }
    int binaryRightSearch(int A[],int left,int right,int target){
        int originalLeft = left, originalRight = right;
        while(left <= right){
            int mid = (left + right)/2;
            if(A[mid] == target){
                if(mid == originalRight || A[mid+1] != target)
                    return mid;
                else
                    left = mid+1;
            }
            else if(A[mid] > target)
                right = mid - 1;
            else
                left = mid + 1;
        }
        return -1;
    }
    vector<int> searchRange(int A[], int n, int target) {
        vector<int> ret(2);
        ret[0] = binaryLeftSearch(A,0,n-1,target);
        ret[1] = binaryRightSearch(A,0,n-1,target);
        return ret;
    }
};