leetcode 34 search for a range

递归

class Solution {public:    typedef vector<int>::size_type sz;    vector<int> searchRange(vector<int>& nums, int target) {        sz size = nums.size();        if (size == 0) return vector<int>(2, -1);        vector<int> ret;        // the ragne we are finding is [left, right]        int left = 0;        int right = size - 1;        int mark = -1;        while(left <= right) {            int piv = (left + right) / 2; //向左偏            if (nums[piv] == target) {                mark = piv; // left mark the index of the number which is equal to target                break;            }            else if (nums[piv] < target){                    left = piv + 1;            }            else {                    right = piv - 1;            }        }        if (mark == -1) // no number in the nums is equal to target             return vector<int>(2, -1);        else {            cout << mark << endl;            ret.push_back(leftFind(nums, 0, mark, target));            ret.push_back(rightFind(nums, mark, size - 1, target));        }        return ret;    }private:    // find the left boundary     int leftFind(vector<int>& v, size_t b, size_t e, int target) {        if (b == e)             return e;        size_t piv = (b + e) / 2;        if (v[piv] == target)             return leftFind(v, b, piv, target);        else            return leftFind(v, piv + 1, e, target);    }    // find the right boundary    int rightFind(vector<int>& v, size_t b, size_t e, int target) {        if (b == e)             return b;        size_t piv = (b + e + 1) / 2;        if (v[piv] == target)            return rightFind(v, piv, e, target);        else            return rightFind(v, b, piv - 1, target);    }};

hehe

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> ret(2, -1);        typedef vector<int>::iterator It;        pair<It, It> p = equal_range(nums.begin(), nums.end(), target);        if (p.first != p.second) {            ret[0] = p.first - nums.begin();            ret[1] = p.second - nums.begin() - 1;        }        return ret;    }};

参考后

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int size = nums.size();        vector<int> ret(2, -1);        if (size == 0) return ret;        int left = -1;        int right = -1;        // find left boundary        int l = 0;        int r = size - 1;        while (l < r) {            int mid = (l + r) / 2; // 向左偏            if (nums[mid] < target)                 l = mid + 1;            else                 r = mid;        }        if (nums[l] == target)             left = l;        else return ret; // 没有数字和 target 相同        // find right boundary        l = left;        r = size - 1;        while (l < r) {            int mid = (l + r + 1) / 2;  // 向右偏 不然 r = mid -1 会小于0            if (nums[mid] > target)                r = mid - 1;            else                 l = mid;  // 不能是 mid -1 因为nums[l] 可能是target        }        right = r;        ret[0] = left;        ret[1] = right;        return ret;     }};