二分查找问题总结

一、最朴素的二分查找问题

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题目

代码

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0, r = nums.size() - 1;
        // 注意：此处是l < =r
        while (l <= r)
        {
            int mid = (l + r) / 2;
            if (nums[mid] == target) return mid;
            else if (nums[mid] < target) l = mid + 1;
            else r = mid - 1;
        }
        return -1;
    }
};

二、找递增数组中x的左右索引值

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题目

代码

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> res;
        // 数组为空，特判一下
        if (!nums.size())
        {
            res = {-1, -1};
            return res;
        }
        // 找左边界
        int l = 0, r = nums.size() - 1;
        while (l < r)
        {
            int mid = (l + r) / 2;
            if (nums[mid] >= target) r = mid;
            else l = mid + 1;
        }
        if (nums[l] != target) res = {-1, -1};
        else
        {
            res.push_back(l);
            // 找右边界
            int l = 0, r = nums.size() - 1;
            while (l < r)
            {
                int mid = (l + r + 1) / 2;
                if (nums[mid] <= target) l = mid;
                else r = mid - 1;
            }
            res.push_back(l);
        }
        return res;
    }
};