一、 缀点成线(LeetCode-5230)
1.1 题目描述
1.2 解题思路
比较简单的一题,利用公式 y = kx + b,确定好k和b就好,并且要考虑一种情况,函数 x = h。
1.3 解题代码
public boolean checkStraightLine(int[][] coordinates) {
int k = Integer.MIN_VALUE;
int x1 = coordinates[0][0];
int y1 = coordinates[0][1];
int x2 = coordinates[1][0];
int y2 = coordinates[1][1];
int dy = y2 - y1;
int dx = x2 - x1;
if (dx != 0) {
k = dy / dx;
}
int b = y1 - k * x1;
boolean onLine = true;
for (int i = 2; i < coordinates.length; i++) {
if (k == Integer.MIN_VALUE) {
if (coordinates[i][0] != x1) {
onLine = false;
break;
}
} else if (coordinates[i][1] != (k * (coordinates[i][0]) + b)) {
onLine = false;
break;
}
}
return onLine;
}
二、 删除子文件夹(LeetCode-5231)
2.1 题目描述
2.2 解题思路
使用两层for循环会导致超时,需要使用字典树解决。
2.3 解题代码
class Solution {
public Tree rootNode = null;
public List<String> removeSubfolders(String[] folder) {
rootNode = new Tree();
rootNode.path = "/";
rootNode.nextMap = new HashMap<>();
List<String> res = new ArrayList<>();
Arrays.sort(folder, (a, b) -> {
return a.length() - b.length();
});
for (int i = 0; i < folder.length; i++) {
if (!isCover(folder[i])) {
res.add(folder[i]);
}
}
return res;
}
public boolean isCover(String path) {
boolean cover = false;
String[] pathArray = path.split("/");
int arrayLen = pathArray.length;
Tree tmp = rootNode;
for (int i = 1; i < arrayLen; i++) {
Tree node = tmp.nextMap.get(pathArray[i]);
if (node == null) {
node = new Tree();
node.path = pathArray[i];
node.nextMap = new HashMap<>();
} else {
if (node.isRoot) {
cover = true;
}
}
if (cover == false && i == arrayLen - 1) {
node.isRoot = true;
}
tmp.nextMap.put(pathArray[i], node);
tmp = node;
}
return cover;
}
}
class Tree {
public String path;
//判断是否是根文件夹
//例如 /a/b 可能是根文件夹,需要在b节点,记录isRoot = true;
public boolean isRoot;
public Map<String, Tree> nextMap;
}