leetcode 34. Search for a Range 及35

leetcode 34题 search for a Range

题意为查找目标值的起始和结束位置，不存在则返回0.

用二分查找即可做到满足条件时间复杂度为O（lgn）.我的思路是首先查找出第一个等于target的值，再在其左，右分别查找。

起初一直超时，原因是while的判断条件和对最左值的更新上，起初直接把最左值（head等）直接赋值为中间值，结果是跳不出循环因为/取小的那个，就会出不了循环。还有对第一次查找后的判断，当正常出循环而不是break 出循环时，需要判断是否是等于target。第一次的循环是为了缩小查找范围。

35题采用34题的第一部分再稍作修改即可

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        vector<int> pos(2,-1);        if (nums.empty() || nums.back() < target)            return pos;        int head,tail,mid;        head = 0;        tail = nums.size() - 1;        if (head == tail)        {            if(target == nums[head])            {                pos[0] = head;                pos[1] = head;            }            return pos;        }                    while(head < tail) //no mid element        {            mid = (head + tail)/2;            if(target == nums[mid])            {               break;                            }            else if (target > nums[mid])            {                head = mid + 1;            }            else                tail = mid;        }        if (head == tail) //只有一个在最后，或者没有找到        {            if (nums[head]== target)            {                pos[0] = head;                pos[1] = head;            }                            return pos;        }        int fistart = mid,fiend = mid,newmids = mid,newmide = mid;        while(head < newmids)        {            if(nums[(head + newmids)/2] < target)            {                                head = (head + newmids)/2 + 1;                            }                        else            {                if(fistart > (head + newmids) / 2)                    fistart = (head + newmids) / 2;                newmids = (head + newmids)/2;            }        }        while(newmide < tail )        {            if(nums[(newmide + tail)/2] > target)          {            tail = (newmide + tail)/2;           }            else            {                if(fiend < (newmide + tail)/2)                    fiend = (newmide + tail)/2;                                           newmide = (tail + newmide)/2 + 1 ;            }        }        if (nums[newmide] == target)            fiend = newmide;                pos[0] = fistart;        pos[1] = fiend;        return pos;                            }       };