[Leetcode] 360. Sort Transformed Array 解题报告
2017-08-14 13:46
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题目:
Given a sorted array of integers nums and integer values a, b and c. Apply a function of the form f(x) = ax2 + bx + c to
each element x in the array.
The returned array must be in sorted order.
Expected time complexity: O(n)
Example:
思路:
这道题目的本质其实是对一个有序数组重新进行排序。由于是一个二次变换,所以分如下情况分别考虑:
1)a == 0:此时函数退化为线性函数,不用排序,直接变换并且返回(如果b < 0的时候需要对结果进行逆转)。
2)a != 0:此时首先计算出二次函数的对称轴(-b / (2 * a)),然后根据距离这个对称轴的远近对nums中的元素进行排序,再进行变换(当然在a < 0的情况下,还需要对变换结果进行逆转)。由于原来的数组已经是有序的了,所以这里的重新排序可以在O(n)的时间内完成。
代码:
class Solution {
public:
vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {
vector<int> ret;
double DOUBLE_MAX = numeric_limits<double>::max();
if(nums.size() == 0) {
return ret;
}
if(a == 0) { // transform directly
for(int i = 0; i < nums.size(); ++i) {
ret.push_back(b * nums[i] + c);
}
if(b < 0) {
reverse(ret.begin(), ret.end());
}
}
else {
double axis = ((double)(-b)) / (2 * a); // re sort accoring to the axis (-b / (2 * a))
int upper = distance(nums.begin(), lower_bound(nums.begin(), nums.end(), axis));
int lower = upper - 1;
while(upper < nums.size() || lower >= 0) {
double upper_differ = (upper == nums.size()) ? DOUBLE_MAX : abs(nums[upper] - axis);
double lower_differ = (lower == -1) ? DOUBLE_MAX : abs(axis - nums[lower]);
if(upper_differ < lower_differ) {
ret.push_back(nums[upper++]);
}
else {
ret.push_back(nums[lower--]);
}
}
for(int i = 0; i < ret.size(); ++i) { // transform after resorting
ret[i] = a * ret[i] * ret[i] + b * ret[i] + c;
}
if(a < 0) {
reverse(ret.begin(), ret.end());
}
}
return ret;
}
};
Given a sorted array of integers nums and integer values a, b and c. Apply a function of the form f(x) = ax2 + bx + c to
each element x in the array.
The returned array must be in sorted order.
Expected time complexity: O(n)
Example:
nums = [-4, -2, 2, 4], a = 1, b = 3, c = 5, Result: [3, 9, 15, 33] nums = [-4, -2, 2, 4], a = -1, b = 3, c = 5 Result: [-23, -5, 1, 7]
思路:
这道题目的本质其实是对一个有序数组重新进行排序。由于是一个二次变换,所以分如下情况分别考虑:
1)a == 0:此时函数退化为线性函数,不用排序,直接变换并且返回(如果b < 0的时候需要对结果进行逆转)。
2)a != 0:此时首先计算出二次函数的对称轴(-b / (2 * a)),然后根据距离这个对称轴的远近对nums中的元素进行排序,再进行变换(当然在a < 0的情况下,还需要对变换结果进行逆转)。由于原来的数组已经是有序的了,所以这里的重新排序可以在O(n)的时间内完成。
代码:
class Solution {
public:
vector<int> sortTransformedArray(vector<int>& nums, int a, int b, int c) {
vector<int> ret;
double DOUBLE_MAX = numeric_limits<double>::max();
if(nums.size() == 0) {
return ret;
}
if(a == 0) { // transform directly
for(int i = 0; i < nums.size(); ++i) {
ret.push_back(b * nums[i] + c);
}
if(b < 0) {
reverse(ret.begin(), ret.end());
}
}
else {
double axis = ((double)(-b)) / (2 * a); // re sort accoring to the axis (-b / (2 * a))
int upper = distance(nums.begin(), lower_bound(nums.begin(), nums.end(), axis));
int lower = upper - 1;
while(upper < nums.size() || lower >= 0) {
double upper_differ = (upper == nums.size()) ? DOUBLE_MAX : abs(nums[upper] - axis);
double lower_differ = (lower == -1) ? DOUBLE_MAX : abs(axis - nums[lower]);
if(upper_differ < lower_differ) {
ret.push_back(nums[upper++]);
}
else {
ret.push_back(nums[lower--]);
}
}
for(int i = 0; i < ret.size(); ++i) { // transform after resorting
ret[i] = a * ret[i] * ret[i] + b * ret[i] + c;
}
if(a < 0) {
reverse(ret.begin(), ret.end());
}
}
return ret;
}
};
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