C++: Square of Leetcode-977 Question Ordered Array
A violent solution, first square and then sort, time complexity O(nlogn)
A double pointer method, time complexity O(n)
Focus on familiarity with the use of double pointers
Article directory
topic
Question:
Given an integer array nums sorted in non-decreasing order, return a new array composed of the square of each number, also sorted in non-decreasing order.
Brute force solution
- First traverse the square and then use sort() to sort, the method is simple, and the time complexity is O(nlogn)
/*
Leetcode-977题
题目：
给你一个按 非递减顺序 排序的整数数组 nums，返回 每个数字的平方 组成的新数组，要求也按 非递减顺序 排序。
思路分析：
1.先平方后，排序
2.先排序后平方，此方法不行，因为元素存在负数，-5的平方是比3平方大的，所以不能先排序
//暴力解法！！！简单题！！
时间复杂度O(n + nlog n)，其实等同于O(nlog n)
*/
#include "iostream"
#include "vector"
#include "algorithm"
using namespace std;
//熟悉vector下的自建函数排序
//自建排序,降序
bool mysort(const int a, const int b)
{
return a > b;
}
//先平方后排序
class Solution
{
private:
public:
vector<int> sortedSquares(vector<int> &nums)
{
for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++)
{
*it = (*it) * (*it);
}
// sort(nums.begin(), nums.end(), mysort);
sort(nums.begin(), nums.end());
return nums;
}
};
//先排序后平方
//会因为元素存在负数而出现排序出错
class Solution2
{
public:
vector<int> sortedSquares(vector<int> &nums)
{
sort(nums.begin(), nums.end());
for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++)
{
*it = (*it) * (*it);
}
return nums;
}
};
int main(int argc, char const *argv[])
{
vector<int> nums;
nums.push_back(-4);
nums.push_back(-1);
nums.push_back(0);
nums.push_back(3);
nums.push_back(10);
Solution s1;
s1.sortedSquares(nums);
//输出
for (vector<int>::iterator it = nums.begin(); it != nums.end(); it++)
{
cout << *it << endl;
}
vector<int> nums2;
nums2.push_back(-4);
nums2.push_back(-1);
nums2.push_back(0);
nums2.push_back(3);
nums2.push_back(10);
Solution2 s2;
s2.sortedSquares(nums2);
cout << "s2先排序后平方" << endl;
//输出
for (vector<int>::iterator it2 = nums2.begin(); it2 != nums2.end(); it2++)
{
cout << *it2 << endl;
}
return 0;
}
double pointer method
- The left and right pointers of the input array are constantly moving, and the pointer to create a new array moves from the tail end to the head end
- Compare the squares of the left and right pointers, the larger pointer moves, the smaller pointer does not move, then get the larger value and move to the left
- Continuously loop to the end of the new array pointer
- The time complexity is O(n)
/*
此题暴力解法简单，但此代码采用巧妙双指针，需灵活使用
思路：
1.输入数组的左右指针不断移动，创建新数组的指针，从尾端还是向头端移动
2.左右指针平方进行比较，较大的指针移动，较小的指针不移动，然后获取较大值再向左移
3.不断循环至新数组指针到头
//时间复杂度为O(n)
*/
#include "iostream"
#include "vector"
using namespace std;
class Solution
{
public:
vector<int> sortedSquares(vector<int> &nums)
{
vector<int> res(nums.size());
int newIndex = nums.size() - 1;
int beginIndex = 0;
int endIndex = nums.size() - 1;
while (newIndex >= 0) //注意范围！
{
//尾指针平方大
//尾指针平方赋值新数组,头指针不动，尾指针左移
if ((nums[endIndex] * nums[endIndex]) >= (nums[beginIndex] * nums[beginIndex]))
{
res[newIndex] = (nums[endIndex] * nums[endIndex]);
endIndex--;
}
//头指针平方大
//头指针平方赋值新数组，尾指针不动，头指针左移
else
{
res[newIndex] = (nums[beginIndex] * nums[beginIndex]);
beginIndex++;
}
//新数组指针左移
newIndex--;
}
return res;
}
};
int main(int argc, char const *argv[])
{
//测试！！！
vector<int> nums;
nums.push_back(-7);
nums.push_back(-3);
nums.push_back(2);
nums.push_back(3);
nums.push_back(10);
Solution s1;
vector<int> res = s1.sortedSquares(nums);
for (vector<int>::iterator it = res.begin(); it != res.end(); it++)
{
cout << *it << endl;
}
return 0;
}
Summarize
This question is a simple question, the focus is on proficient use of double pointers, knowing when to use double pointers is more efficient, and what the meanings of double pointers
represent
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