在数组中找到和为定值的两个数,返回数组下标。
这个题目就是一个排序的问题,但是不能改变数组,用快排的方式存储数组的索引值即可,上代码。
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <vector>
using namespace std;
void quicksort_index(int *arr, int *index, int low, int high)
{
if (low >= high)
return;
int i = low;
int j = high;
int tmp = index[j];
int val = arr[index[j]];
while (i < j)
{
while (i < j && arr[index[i]] <= val)
i++;
index[j] = index[i];
while (i < j && arr[index[j]] > val)
j--;
index[i] = index[j];
}
index[i] = tmp;
quicksort_index(arr, index, low, i-1);
quicksort_index(arr, index, i+1, high);
}
void get_two_sum_index(int *arr, int *index_arr, int len, int two_sum, std::vector<int> &ivec)
{
int start_index = 0;
int end_index = len;
int i = 0;
int j = 0;
ivec.clear();
while (start_index < end_index)
{
i = index_arr[start_index];
j = index_arr[end_index];
if (two_sum == arr[i] + arr[j]) {
ivec.push_back(i);
ivec.push_back(j);
start_index++;
end_index--;
} else if (two_sum > arr[i] + arr[j]) {
start_index++;
} else {
end_index--;
}
}
}
int main(int argc, char *argv[])
{
//int arr[] = {1, -2, 3, 10, -4, 7, 2, -5};
int arr[] = {102, 21, 39, 210, 94, 72, 2, 5, 167};
int len = sizeof(arr) / sizeof(int);
std::cout << "array data" << std::endl;
int *index_arr = new int[len];
int i = 0;
for (i = 0; i < len; i++)
{
std::cout << arr[i] << " ";
index_arr[i] = i;
}
std::cout << std::endl;
quicksort_index(arr, index_arr, 0, len-1);
for (i = 0; i < len; i++)
{
std::cout << arr[i] << " ";
}
std::cout << std::endl;
for (i = 0; i < len; i++)
{
std::cout << arr[index_arr[i]] << " ";
}
std::cout << std::endl;
int two_sum = 111;
std::cout << "to get two sum: " << two_sum << std::endl;
std::vector<int> ivec;
get_two_sum_index(arr, index_arr, len-1, two_sum, ivec);
std::vector<int>::iterator iter;
int min_val;
int max_val;
for (iter = ivec.begin(); iter != ivec.end(); iter++)
{
std::cout << "[min_index, max_index]: " << *iter << ", ";
min_val = arr[*iter];
iter++;
max_val = arr[*iter];
std::cout << *iter << std::endl;
std::cout << "[min_val, max_val]: " << min_val << " " << max_val << std::endl;
}
return 0;
}