LintCode 280: Closest City
280. Closest City
There are many cities on a two-dimensional plane, all cities have their own names c[i]c[i], and location coordinates (x[i],y[i])(x[i],y[i]) (both are integers). There are qq group queries, and each group query gives a city name. You need to answer the city name that is the closest to the city and has the same xx or yy.
Example
Example 1：
Input: x = [3, 2, 1] y = [3, 2, 3] c = ["c1", "c2", "c3"] q = ["c1", "c2", "c3"]
Output: ["c3", "NONE", "c1"]
Explanation: For c1, c3 is the same as its y, the closest distance is (3-1)+(3-3)=2
For c2, no city is the same as its x or y
For c3, c1 is the same as his y, the closest distance is (3-1)+(3-3)=2
Notice
If there are multiple answers that satisfy the conditions, the one with the smallest lexicographic name is output.
The distance here is the Euler distance: the absolute value of the coordinate difference of xx plus the absolute value of the coordinate difference of yy.
0 ≤ Number of cities ≤ 10^5,
0 ≤ Ask for the number of groups ≤ 10,
1 ≤ Coordinate range 10^9
解法1：
这题被标识为简单，但我觉得其实做出来挺花时间。
我的解法是把每个X和Y坐标都建一个vector数组，装在这个X或Y坐标上的城市列表。然后排序。对于query数组，找到这个城市在对应X和Y坐标上的位置，然后前后找对应的邻近城市，然后比较大小。
注意：
1) for (auto & elem : x_cities) 和 for (auto & elem : y_cities) 要加引用，因为内部x_cities和y_cities排序，产生了变化。
2) 当比较大小是相等时，用字典排序。
这个解法时间复杂度是O(nlogn)，效率不高。
class Solution {
public:
/**
* @param x: an array of integers, the x coordinates of each city[i]
* @param y: an array of integers, the y coordinates of each city[i]
* @param c: an array of strings that represent the names of each city[i]
* @param q: an array of strings that represent the names of query locations
* @return: the closest city for each query
*/
vector NearestNeighbor(vector &x, vector &y, vector &c, vector &q) {
int n = x.size();
vector result;
unordered_map>> x_cities; //x->list of (y, city)
unordered_map>> y_cities; //y->list of (x, city)
unordered_map city_x; //city->x
unordered_map city_y; //city->y
for (int i = 0; i < n; ++i) {
x_cities[x[i]].push_back({y[i], c[i]});
city_x[c[i]] = x[i];
}
for (int i = 0; i < n; ++i) {
y_cities[y[i]].push_back({x[i], c[i]});
city_y[c[i]] = y[i];
}
for (auto &elem : x_cities) { //注意加引用
sort(elem.second.begin(), elem.second.end()); //sort y values under the same x
}
for (auto &elem : y_cities) { //注意加引用
sort(elem.second.begin(), elem.second.end()); //sort x values under the same y
}
for (int i = 0; i < q.size(); ++i) {
int min_dist = INT_MAX;
string closest_city = "NONE";
int target_x = city_x[q[i]];
int target_y = city_y[q[i]];
int dist = 0;
if (x_cities[target_x].size() > 1) {
int find_y_index = binary_search(x_cities[target_x], target_y);
if (find_y_index != -1) { //one city does not need to search
if (find_y_index > 0) {
dist = x_cities[target_x][find_y_index].first - x_cities[target_x][find_y_index - 1].first;
if (min_dist > dist || (min_dist == dist && x_cities[target_x][find_y_index - 1].second < closest_city)) {
min_dist = dist;
closest_city = x_cities[target_x][find_y_index - 1].second;
}
}
if (find_y_index < x_cities[target_x].size() - 1) {
dist = x_cities[target_x][find_y_index + 1].first - x_cities[target_x][find_y_index].first;
if (min_dist > dist || (min_dist == dist && x_cities[target_x][find_y_index + 1].second < closest_city)) {
min_dist = dist;
closest_city = x_cities[target_x][find_y_index + 1].second;
}
}
}
}
if (y_cities[target_y].size() > 1) { //one city does not need to search
int find_x_index = binary_search(y_cities[target_y], target_x);
if (find_x_index != -1) {
if (find_x_index > 0) {
dist = y_cities[target_y][find_x_index].first - y_cities[target_y][find_x_index - 1].first;
if (min_dist > dist || (min_dist == dist && y_cities[target_y][find_x_index - 1].second < closest_city)) {
min_dist = dist;
closest_city = y_cities[target_y][find_x_index - 1].second;
}
}
if (find_x_index < y_cities[target_y].size() - 1) {
dist = y_cities[target_y][find_x_index + 1].first - y_cities[target_y][find_x_index].first;
if (min_dist > dist || min_dist == dist && y_cities[target_y][find_x_index + 1].second < closest_city) {
min_dist = dist;
closest_city = y_cities[target_y][find_x_index + 1].second;
}
}
}
}
result.push_back(closest_city);
}
return result;
}
private:
int binary_search(vector> arr, int target) {
int start = 0, end = arr.size() - 1;
while(start + 1 < end) {
int mid = start + (end - start) / 2;
if (arr[mid].first < target) {
start = mid;
} else if (arr[mid].first > target) {
end = mid;
} else {
return mid;
}
}
if (arr[end].first == target) return end;
if (arr[start].first == target) return start;
return -1;
}
}; 解法2：直接硬比。时间复杂度O(mn)。
class Solution {
public:
/**
* @param x: an array of integers, the x coordinates of each city[i]
* @param y: an array of integers, the y coordinates of each city[i]
* @param c: an array of strings that represent the names of each city[i]
* @param q: an array of strings that represent the names of query locations
* @return: the closest city for each query
*/
vector NearestNeighbor(vector &x, vector &y, vector &c, vector &q) {
int cx, cy;
vector near;
for(int queryIndex = 0; queryIndex < q.size(); queryIndex++)
{
string city = "NONE";
int minDistance = 0;
int findIndex = 0;
for(int i = 0; i < c.size(); i++)
{
if(c[i] == q[queryIndex])
{
cx = x[i];
cy = y[i];
findIndex = i;
break;
}
}
for(int i = 0; i < c.size(); i++)
{
if(i == findIndex)
{
continue;
}
if(x[i] == cx)
{
if(abs(cy - y[i]) < minDistance || minDistance == 0)
{
minDistance = abs(cy - y[i]);
city = c[i];
}
if(abs(cy - y[i]) == minDistance && c[i] < city)
{
minDistance = abs(cy - y[i]);
city = c[i];
}
}
if(y[i] == cy)
{
if(abs(cx - x[i]) < minDistance || minDistance == 0)
{
minDistance = abs(cx - x[i]);
city = c[i];
}
if(abs(cx - x[i]) == minDistance && c[i] < city)
{
minDistance = abs(cx - x[i]);
city = c[i];
}
}
}
near.push_back(city);
}
return near;
}
};