【LeetCode算法练习（C++）】Search for a Range

Stella-Song

于 2017-11-27 09:39:39 发布

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分类专栏：算法分析文章标签： leetcode c++

本文链接：https://blog.youkuaiyun.com/sst2230879/article/details/78641885

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这篇博客介绍了一个LeetCode算法题，题目要求在升序排列的整数数组中找到给定目标值的起始和结束位置。博主提供了解决方案，采用两次二分查找以达到O(log n)的时间复杂度，对于目标值不存在的情况返回[-1, -1]。在给定的示例中，如数组[5, 7, 7, 8, 8, 10]和目标值8，返回的结果是[3, 4]。" 78811346,7418095,STM32Cube生成STM32F103C8T6 USB CDC驱动教程,"['STM32', 'C语言', 'USB通信']

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题目：
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].

链接：Search for a Range
解法：两次二分查找寻找数字的两个边界。时间O(logn)


class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> ans;
        int l = 0, r = nums.size() - 1;
        if (r < 0) {
            ans.push_back(-1);
            ans.push_back(-1);
            return ans;
        }
        if (target < nums[l] || target > nums[r]) {
            ans.push_back(-1);
            ans.push_back(-1);
            return ans;
        }
        while (l < r) {
            int m = (l + r) / 2;
            if (nums[m] < target && nums[m + 1] >= target) {
                r = m + 1;
                break;
            }
            else if (nums[m] < target) l = m;
            else r = m;
        }
        ans.push_back(r);
        l = 0, r = nums.size() - 1;
        while (l < r) {
            int m = (l + r) / 2;
            if (nums[m] <= target && (nums[m + 1] > target || m == nums.size() - 1)) {
                l = m;
                break;
            }
            else if (nums[m] <= target) l = m + 1;
            else r = m;
        }
        ans.push_back(l);
        if (ans[0] > ans[1]) ans[0] = ans[1] = -1;
        return ans;
    }
};

Runtime: 9 ms