Sword Finger Offer 53-I. Find the number I in the sorted array (binary search variant, that is, lower_bound(), upper_bound() by hand)

Thursday, February 4, 2021, the weather is fine [Don’t lament the past, don’t waste the present, don’t fear the future]

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1. Introduction

Sword Finger Offer 53-I. Find the number I in the sorted array

2. Solution

2.1 lower_bound 结合 upper_bound

lower_bound (): returns the first is greater than or equal to a target element location, it is less than the return end ()
upper_bound, (): returns the first is greater than the target positions of the elements are less than or equal, then return end ()

class Solution {
    
    
public:
    int search(vector<int>& nums, int target) {
    
    
        vector<int>::iterator it_beg = lower_bound(nums.begin(),nums.end(),target);
        // target不存在的情况
        if(it_beg==nums.end() || *it_beg != target) return 0;
        // 因为上面已经判断过target不存在的情况，所以这里要么返回end()，要么返回第一个
        // 大于target的元素位置，这两种情况可以合并
        vector<int>::iterator it_end = upper_bound(nums.begin(),nums.end(),target);
        // 返回最终结果
        return it_end - it_beg;   
    }
};

2.2 手撸 lower_bound()、upper_bound()

Here is the variant of binary search, lower_bound() and upper_bound() by hand.
The following is the code. There are 4 details in total. It is recommended to memorize it so as not to be nervous during the interview and forget whether it should be =. Should it be written as -1 or plus +1.

class Solution {
    
    
public:
    int search(vector<int>& nums, int target) {
    
    
        if(nums.empty()) return 0;
        int idx_beg = 0, idx_end = 0;
        int n = nums.size();

        // 返回 第一个等于target 的元素位置（类似lower_bound，稍有改动），都不等于的话返回 0
        int l = 0, r = n - 1; // 细节1：r = n - 1
        while(l < r){
    
     // 细节2：l < r
            int m = l + (r - l)/2;
            // 细节3：三种情况分开写，便于理解
            if(nums[m] < target) l = m + 1;
            else if(nums[m] == target) r = m;
            else  r = m - 1;
        }
        // 细节4：l == r 的情况单独判断
        if(nums[l]==target) idx_beg = l;
        else return 0;

        // 返回 第一个大于target 的元素位置（类似upper_bound，稍有改动），都小于等于的话
        // 返回最后的索引+1（其实就是对应end()）
        l = 0, r = n - 1; // 细节1：r = n - 1
        while(l < r){
    
     // 细节2：l < r
            int m = l + (r - l)/2;
            // 细节3：三种情况分开写，便于理解
            if(nums[m] < target) l = m + 1;
            else if(nums[m] == target) l = m + 1;
            else  r = m;
        }
        // 细节4：l == r 的情况单独判断
        // 如果这里等于，那么就都小于等于target，返回最后的索引+1（对应end()）
        if(nums[l]==target) idx_end = l + 1;
        // 否则，返回第一个大于target的元素位置
        else idx_end = l;
    
        return idx_end - idx_beg;
    }
};

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references

"Sword Finger Offer Second Edition"

https://www.cnblogs.com/mxj961116/p/11945444.html