LeetCode】34. 搜索范围 结题报告 (C++)

原题地址:https://leetcode-cn.com/problems/search-for-a-range/description/

题目描述:

给定一个按照升序排列的整数数组 nums,和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值,返回 [-1, -1]

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]

示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: [-1,-1]

解题方法:

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int i = 0;
        int j = nums.size()-1;
        int mid;
        vector<int> ans;
        while(i <= j){
            mid = (i + j)/2;
            if(nums[mid] == target){
                int m1 = (i + mid)/2;
                int m2 = (j + mid)/2 + 1;
                int tmp = mid;
                while(i != tmp){
                    if(nums[m1] != target)
                        i = m1 + 1;
                    else
                        tmp = m1;
                    m1 = (i + tmp)/2;
                    
                }
                tmp = mid;
                while(tmp != j){
                    if(nums[m2] != target)
                        j = m2 - 1;
                    else
                        tmp = m2;
                    m2 = (j + tmp)/2 + 1;
                }
                ans.push_back(i);
                ans.push_back(j);
                return ans;
            }
            else if(nums[mid] > target)
                j = mid - 1;
            else
                i = mid + 1;
        }
        ans.push_back(-1);
        ans.push_back(-1);
        return ans;
    }
};

没什么技巧,二分法一直怼就行了。。。。

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转载自blog.csdn.net/qq_32805671/article/details/80200903