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;    }};