Description
You are given an array trees where trees[i] = [xi, yi] represents the location of a tree in the garden.
Fence the entire garden using the minimum length of rope, as it is expensive. The garden is well-fenced only if all the trees are enclosed.
Return the coordinates of trees that are exactly located on the fence perimeter. You may return the answer in any order.
Example 1:
Input: trees = [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]]
Output: [[1,1],[2,0],[4,2],[3,3],[2,4]]
Explanation: All the trees will be on the perimeter of the fence except the tree at [2, 2], which will be inside the fence.
Example 2:
Input: trees = [[1,2],[2,2],[4,2]]
Output: [[4,2],[2,2],[1,2]]
Explanation: The fence forms a line that passes through all the trees.
Constraints:
1 <= trees.length <= 3000trees[i].length == 20 <= xi, yi <= 100- All the given positions are unique.
Solutions
Solution 1
PythonJavaC++Go
class Solution:
def outerTrees(self, trees: List[List[int]]) -> List[List[int]]:
def cross(i, j, k):
a, b, c = trees[i], trees[j], trees[k]
return (b[0] - a[0]) * (c[1] - b[1]) - (b[1] - a[1]) * (c[0] - b[0])
n = len(trees)
if n < 4:
return trees
trees.sort()
vis = [False] * n
stk = [0]
for i in range(1, n):
while len(stk) > 1 and cross(stk[-2], stk[-1], i) < 0:
vis[stk.pop()] = False
vis[i] = True
stk.append(i)
m = len(stk)
for i in range(n - 2, -1, -1):
if vis[i]:
continue
while len(stk) > m and cross(stk[-2], stk[-1], i) < 0:
stk.pop()
stk.append(i)
stk.pop()
return [trees[i] for i in stk](code-box)
class Solution {
public int[][] outerTrees(int[][] trees) {
int n = trees.length;
if (n < 4) {
return trees;
}
Arrays.sort(trees, (a, b) -> { return a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]; });
boolean[] vis = new boolean[n];
int[] stk = new int[n + 10];
int cnt = 1;
for (int i = 1; i < n; ++i) {
while (cnt > 1 && cross(trees[stk[cnt - 1]], trees[stk[cnt - 2]], trees[i]) < 0) {
vis[stk[--cnt]] = false;
}
vis[i] = true;
stk[cnt++] = i;
}
int m = cnt;
for (int i = n - 1; i >= 0; --i) {
if (vis[i]) {
continue;
}
while (cnt > m && cross(trees[stk[cnt - 1]], trees[stk[cnt - 2]], trees[i]) < 0) {
--cnt;
}
stk[cnt++] = i;
}
int[][] ans = new int[cnt - 1][2];
for (int i = 0; i < ans.length; ++i) {
ans[i] = trees[stk[i]];
}
return ans;
}
private int cross(int[] a, int[] b, int[] c) {
return (b[0] - a[0]) * (c[1] - b[1]) - (b[1] - a[1]) * (c[0] - b[0]);
}
}(code-box)
class Solution {
public:
vector<vector<int>> outerTrees(vector<vector<int>>& trees) {
int n = trees.size();
if (n < 4) return trees;
sort(trees.begin(), trees.end());
vector<int> vis(n);
vector<int> stk(n + 10);
int cnt = 1;
for (int i = 1; i < n; ++i) {
while (cnt > 1 && cross(trees[stk[cnt - 1]], trees[stk[cnt - 2]], trees[i]) < 0) vis[stk[--cnt]] = false;
vis[i] = true;
stk[cnt++] = i;
}
int m = cnt;
for (int i = n - 1; i >= 0; --i) {
if (vis[i]) continue;
while (cnt > m && cross(trees[stk[cnt - 1]], trees[stk[cnt - 2]], trees[i]) < 0) --cnt;
stk[cnt++] = i;
}
vector<vector<int>> ans;
for (int i = 0; i < cnt - 1; ++i) ans.push_back(trees[stk[i]]);
return ans;
}
int cross(vector<int>& a, vector<int>& b, vector<int>& c) {
return (b[0] - a[0]) * (c[1] - b[1]) - (b[1] - a[1]) * (c[0] - b[0]);
}
};(code-box)
func outerTrees(trees [][]int) [][]int {
n := len(trees)
if n < 4 {
return trees
}
sort.Slice(trees, func(i, j int) bool {
if trees[i][0] == trees[j][0] {
return trees[i][1] < trees[j][1]
}
return trees[i][0] < trees[j][0]
})
cross := func(i, j, k int) int {
a, b, c := trees[i], trees[j], trees[k]
return (b[0]-a[0])*(c[1]-b[1]) - (b[1]-a[1])*(c[0]-b[0])
}
vis := make([]bool, n)
stk := []int{0}
for i := 1; i < n; i++ {
for len(stk) > 1 && cross(stk[len(stk)-1], stk[len(stk)-2], i) < 0 {
vis[stk[len(stk)-1]] = false
stk = stk[:len(stk)-1]
}
vis[i] = true
stk = append(stk, i)
}
m := len(stk)
for i := n - 1; i >= 0; i-- {
if vis[i] {
continue
}
for len(stk) > m && cross(stk[len(stk)-1], stk[len(stk)-2], i) < 0 {
stk = stk[:len(stk)-1]
}
stk = append(stk, i)
}
var ans [][]int
for i := 0; i < len(stk)-1; i++ {
ans = append(ans, trees[stk[i]])
}
return ans
}(code-box)
