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Building Outline.cpp
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Building Outline.cpp
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/*
Given N buildings in a x-axis,each building is a rectangle and can be represented by a triple (start, end, height),
where start is the start position on x-axis, end is the end position on x-axis and height is the height of the building.
Buildings may overlap if you see them from far away,find the outline of them。An outline can be represented by a triple,
(start, end, height), where start is the start position on x-axis of the outline, end is the end position on x-axis and
height is the height of the outline.
Link: http://www.lintcode.com/en/problem/building-outline/
Example: Given 3 buildings:
[
[1, 3, 3],
[2, 4, 4],
[5, 6, 1]
]
The outlines are:
[
[1, 2, 3],
[2, 4, 4],
[5, 6, 1]
]
Solution: Use priority_queue to solve this problm.
Source: None
*/
class Solution {
public:
/**
* @param buildings: A list of lists of integers
* @return: Find the outline of those buildings
*/
vector<vector<int> > buildingOutline(vector<vector<int> > &buildings) {
// write your code here
vector<vector<int> > res;
const int size = 300000;
int n = buildings.size();
if(n == 0) {
return res;
}
sort(buildings.begin(), buildings.end());
priority_queue<vector<int>> hp;
int bp = 0, cur_h = -1, start = -1;
vector<int> build = buildings[bp];
for(int i = 0; i < size; i++){
while(!hp.empty()){
if(hp.top()[2] < i) {
hp.pop();//height overflow
} else {
break;
}
}
if(!hp.empty()) {
if(hp.top()[0] == cur_h) {
res.back()[1] = i;
} else {
cur_h = hp.top()[0];
res.push_back({start,i,cur_h});
}
}
while(bp < n && build[0] == i) {
hp.push({build[2],build[0],build[1]});
if(++bp < n) {
build = buildings[bp];
}
}
start = i;
}
return res;
}
};