LeetCode 34. Search for a Range

Description

Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

my program

(1)时间复杂度为O（n）

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {        int i = 0, j = nums.size() -1;        vector<int> res;        while (i < nums.size())        {            if (nums[i] == target)                  break;            i++;        }        while ( j>= 0)        {            if (nums[j] == target)                  break;            j--;        }        if (i>=nums.size())        {            res.push_back(-1);            res.push_back(-1);        }else        {            res.push_back(i);            res.push_back(j);        }        return res;    }};

(2)时间复杂度为O（logn），利用二分查找

class Solution {public:    vector<int> searchRange(vector<int>& nums, int target) {    int begin = 0, end = nums.size(), mid, left, right;    while (begin < end) {        mid = (begin + end) / 2;        if (nums[mid] >= target)            end = mid;        else            begin = mid + 1;    }    left = begin;    begin = 0, end = nums.size();    while (begin < end) {        mid = (begin + end) / 2;        if (nums[mid] > target)            end = mid;        else            begin = mid + 1;    }    right = begin;    return left == right ? vector<int> {-1,-1} : vector<int> {left,right-1};    }};

先找左边界。当mid >= target,将end移动到mid，否则（mid < target），begin = mid+1；

再找右边界。 当mid > target,将end移动到mid，否则（mid <= target），begin = mid+1；最后begin和end可能在target的左面一个。

如果没有数组中target的话，那么left和right会指到同一个值上，否则的话就意味着出现了target，此时left指向第一个出现的target，而right是指向最后一个target的下一个值，所以right需要减去1.