如何在不使用标准算法的情况下将c元素添加到排序向量中？

Question

My code does not work in some cases: when c1 is 4 , it doesn't output anything, but it works for numbers greater than 9. Why is that?

#include<iostream>
#include<vector>
using namespace std;
void insertfast(vector<int>&v, int c)
{
if (c >= v[v.size()-1])v.push_back(c);
if (c <= v[0])v.insert(v.begin(), c);
int min = 1;
int max = v.size();
while (v.size() != 9) {
    int i = (min + max) / 2;
    if (v[i - 1] <= c && c <= v[i])v.insert(v.begin() + i, c);
    if (v[i] <= c && c <= v[i + 1])v.insert(v.begin() + (i + 1), c);
    if (c < v[i])
        max = i;
    else
        min = i;
}
}
int main()
{
vector<int>v1 = { 2,5,9,22,44,55,88,777 };
int c1 = 4;

insertfast(v1, c1);

for (int i = 0; i < 9; i++)
    cout << v1[i] << endl;
}

Answer 1

You have two logic problems in your code:

1. For your test data in insertfast() function you step into while() , when you step into first if (v[i - 1] <= c && c <= v[i]) you insert 4 value and in becomes v[1] , and you have step right after that into next if (v[i] <= c && c <= v[i + 1]) , so you need to set second if as else if , like this:

    if (v[i - 1] <= c && c <= v[i]) { v.insert(v.begin() + i, c); } else if (v[i] <= c && c <= v[i + 1]) { v.insert(v.begin() + (i + 1), c); } 

2. From the previous item your vector increased for 2 elements, so his size becomes 10 , and you had infinite loop while (v.size() != 9) , I think you can just do break to leave loop in case of success insert:

    if (v[i - 1] <= c && c <= v[i]) { v.insert(v.begin() + i, c); break; } else if (v[i] <= c && c <= v[i + 1]) { v.insert(v.begin() + (i + 1), c); break; } 

There is no need to run loop further, if you have inserted element. Actually 2nd item with break will fix and else missing problem, in case of insertion it will leave loop and won't be able to step into second if.

Answer 2

Let's just take a second to logically step through the code here, since I assume this is a homework question, it's important to understand how to do this. There are tools like debuggers that are used on larger projects and more difficult problems, but small logic errors like this happen all the time.

Ok, so insertfast() is a function to insert an int into a vector<int> in an ordered manner. The first thing you do is compare against the first and last elements of the vector.

if (c >= v[v.size() - 1]) 
    v.push_back(c); 
if (c <= v[0]) 
    v.insert(v.begin(), c);

We can only be one of these things, and if either are true we are done. So the best option here is to just return from the function, we don't care about doing anything else. We know that, but the program doesn't. It'll keep on comparing things we don't want to compare and bug out.

if (c >= v[v.size() - 1]) {
    v.push_back(c);
    return;
}
if (c <= v[0]) {
    v.insert(v.begin(), c);
    return;
}

What if both of these statements are false, and we are still in the function? We still need to figure where to insert this value, so we loop through the elements.

while (v.size() != 9)

This is a very poor conditional loop, since it will only work if we pass a vector of size 8 into the function. With your particular arithmetic, you essentially want to loop until some value is added to the vector. I recommend an unconditional loop (we are trying to be "fast" after all) aka an infinite loop, and explicitly return or break within the loop.

int min = 1;
int max = v.size();
while (1) {
    int i = (min + max) / 2;
    if (v[i - 1] <= c && c <= v[i]) {
        v.insert(v.begin() + i, c);
        return;
    }
    if (v[i] <= c && c <= v[i + 1]) {
        v.insert(v.begin() + (i + 1), c);
        return;
    }
    if (c < v[i])
        max = i;
    else
        min = i;
}

break instead of return will also work just fine here, since if you break out of this loop, the end of the function is hit and you just return anyways. So, we solved the problem just by rethinking the design and stepping through the code.