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 // Find the right border ： The left side conforms to  →  For maximum 
    const check = (m)=>{
    } // Coincidence function 
    let [l,r] = [0,n-1] // Left and right 
    while(l < r){
    
        const m = Math.ceil((l+r)/2)  // Rounding up , Prevent rounding down l = m It's stuck 
        // Meet the requirements ： All on the left meet the requirements 
        if(check(m)) l = m
        // Does not meet the requirements ： All on the right do not meet the requirements 
        else r = m-1 
    } 

    if(check(l)) return l // Already found , You can add sentinel elements on the left , Make sure that the element can be found 
    else return -1// Can't find

here l from 0 Start , In fact, it may not be found , So can I output l Made a check Judge , If false, In fact, it was not found , Not found 0 This position , So we need to return -1;

Of course, there is another way to start l = -1 Start looking for , Finally, there is no need to judge check, Direct judgment l == -1.

275. H Index II
[222. The number of nodes in a complete binary tree ](https://leetcode-cn.com/problems/count-
1292. The maximum side length of an element and a square less than or equal to the threshold complete-tree-nodes/)

minimum value

Looking for the left boundary , That is, the right side is consistent with the answer → Use a bisection template lower_bound

 // Find the left border ： The right side conforms to  →  For the minimum 
    const check = (m)=>{
    } // Coincidence function 
    let [l,r] = [ Left , Right ] // Left and right 
    while(l < r){
    
        const m = Math.floor((l+r)/2) // Rounding down , Rounding up will be in r = m It's stuck 
        // Meet the requirements ： All on the right meet the requirements 
        if(check(m)) r = m
        // Does not meet the requirements ： All on the left do not meet the requirements 
        else l = m+1 
    } 
    if(check(l)) return l // Already found , You can add sentinel elements on the right , Make sure that the element can be found 
    else return -1 // Can't find

436. Look for the right range

（ Not limited to left and right pointers , It's just a question of dichotomy ）
1608. The eigenvalues of a particular array
1898. The maximum number of characters that can be removed
1870. The minimum speed of a train arriving on time Decimal conversion , Division conversion is very useful
1283. The minimum divisor that makes the result not exceed the threshold
1712. The number of schemes that divide an array into three subarrays
1574. Delete the shortest subarray to make the remaining arrays ordered Personally, I feel the problem in dichotomy …

Pile up

I have made some summaries of many problems ：
Heap & Big root pile ：
Xiaogen == Big top == less< T > == return a < b; == Small priorities are high == There are smaller numbers in the heap, and the smaller the number, the more at the bottom of the heap == top What comes out is the top of the pile , Is the largest of the smaller values == pop What comes out is a larger value
Collocation top() pop(),, It is precisely because each time the priority is small , And throw out the ones with low priority

703. The... In the data stream K Big element
Interview questions 17.09. The first k Number
347. front K High frequency elements
973. Closest to the origin K A little bit
1337. The weakest in the matrix K That's ok
priority_queue q(greater<pair<int, int>>(), move(power)); Right quoting , strong
373. Look for the smallest K Pairs of numbers More uncomfortable question , Although it looks similar to the above two About pair There are answers to how to customize the comparison method
786. The first K The smallest prime fraction This question and the one above can experience customization pair How to write
778. Swimming in a rising pool
295. Median of data flow Two piles
1606. Find the server that handles the most requests