[leetcode] 34. Search for a Range解題レポート

タイトルリンク:https://leetcode.com/problems/search-for-a-range/
Given a sorted array of integers, 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] .
考え方:二分検索で左右の境界を見つければいい.左右の境界を見つければ判断条件を変えればいい.
コードは次のとおりです.

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int left = 0, len = nums.size(), right = len - 1, index1=-1,index2 = -1;
        vector<int> result(2, -1);
        while(left <= right)//        
        {
            int mid = (left+right)/2;
            if(nums[mid] < target)
                left = mid +1;
            else
                right = mid -1;
        }
        index1 = left;
        right = len -1;
        while(left <= right)//         targe   
        {
            int mid = (left+right)/2;
            if(nums[mid] <= target)
                left = mid + 1;
            else
                right = mid -1;
        }
        index2 = left -1;
        if(nums[index1] != target || nums[index2] != target)//           targe，       
            return result;
        result[0] = index1;
        result[1] = index2;
        return result;
    }
};