8.16 56. Merge Intervals
Given an array of intervals where intervalsi = starti, endi, merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
- 合并区间
本题也是重叠区间问题，如果昨天三道都吸收的话，本题就容易理解了。
https://leetcode.cn/problems/merge-intervals/description/
https://programmercarl.com/0056.合并区间.html
返回值是长度不一定的二维数组？不要直接new二维数组，先将一维数组存入list，再toArray即可。
这里是合并区间，需要将左右边界都保存下来。
```java
public class mergeIntervals {
public int\[\]\[\] merge(int\[\]\[\] intervals) {
List<int\[\]> result = new ArrayList<>();
Arrays.sort(intervals,(a,b) -> Integer.compare(a0,b0));
int left = intervals00;
int right = intervals01;
for (int i = 1; i < intervals.length; i++) {
//这里其实是将当前左边界和上一组的右边界进行比对，并更新右边界。
if(intervalsi0 <= right){
right = Math.max(intervalsi1,right);
}else{
result.add(new int\[\]{left,right});
left = intervalsi0;
right = intervalsi1;
}
}
result.add(new int\[\]{left,right});
return result.toArray(new intresult.size()\[\]);
}
}
class mergeIntervalsTest {
public static void main(String\[\] args) {
mergeIntervals example = new mergeIntervals();
int\[\]\[\] intervals = {
//{1,3},{2,6},{8,10},{15,18}
{2,3},{4,5},{6,7},{8,9},{1,10}
//{-2147483646,-2147483645},{2147483646,2147483647}
};
int\[\]\[\] result = example.merge(intervals);
for(int\[\] line : result){
System.out.print("[ ");
for(int i : line){
System.out.print(i + " ,");
}
System.out.print(" ] ");
}
}
}
```
8.17 738. Monotone Increasing Digits
An integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.
Given an integer n, return the largest number that is less than or equal to n with monotone increasing digits.
https://leetcode.cn/problems/monotone-increasing-digits/description/
https://programmercarl.com/0738.单调递增的数字.html
这题的巧妙之处在于，找到第一个非单调递增的数字后，将它-1，后面的数字都改为9.
```java
public class monotoneIncreasingDigits {
public monotoneIncreasingDigits() { }
public int monotoneIncreasingDigits(int n) {
String s = String.valueOf(n);
char\[\] chars = s.toCharArray();
//用于记录末尾的9从哪里开始
int start = chars.length;
for (int i = chars.length - 1; i > 0; i--) {
if(charsi-1 > charsi){
charsi-1--;
start = i;
}
}
// for (int i = chars.length - 2; i >= 0; i--) {
// if(charsi+1 < charsi){
// charsi--;
// start = i + 1;
// }
// } //没有区别，写法不一样
// System.out.println("start: " + start);
for (int i = start; i < chars.length; i++) {
charsi = '9';
}
return Integer.parseInt(new String(chars));
}
}
class monotoneIncreasingDigitsTest {
public static void main(String\[\] args) {
monotoneIncreasingDigits example = new monotoneIncreasingDigits();
System.out.println(example.monotoneIncreasingDigits(3452331));
}
}
```
8.18 968. Binary Tree Cameras（选做）
You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.
Return the minimum number of cameras needed to monitor all nodes of the tree.
https://leetcode.cn/problems/binary-tree-cameras/description/
https://programmercarl.com/0968.监控二叉树.html
碰到叶子节点，就让它的父节点放一个摄像头；向上遍历的时候每隔两个节点放一个摄像头。
如何实现？通过记录状态
三个状态：
0.无覆盖（没有被摄像头覆盖）
1.有摄像头
2.有覆盖
无摄像头的状态可以被0和1替代。
目的：尽可能地让叶子的父节点放摄像头。
三种情况：
1.左右孩子都有覆盖（2），父节点则无覆盖。（叶子节点返回2）
2.左右孩子有一个无覆盖，父节点需要装摄像头
3.左右孩子有一个有摄像头，父节点有覆盖。
4.遍历结束后，头结点没有覆盖，需要加一个摄像头
```java
public class binaryTreeCameras {
int result = 0;
public int minCameraCover(TreeNode root) {
//当头节点没有被覆盖，需要将一个头布置在头结点上。
if(traversal(root) == 0) result++;
return result;
}
public int traversal(TreeNode root){
//根节点不布置摄像头，直接返回有覆盖
if(root == null){
return 2;
}
int left = traversal(root.left);
int right = traversal(root.right);
// 如果左右节点都覆盖了的话, 那么本节点的状态就应该是无覆盖,没有摄像头
if(left == 2 && right == 2){
return 0;
//左右节点存在无覆盖状态,那 根节点此时应该放一个摄像头
}else if(left==0||right==0){
result++;
return 1;
}else{
//左右节点至少存在一个摄像头
return 2;
}
}
}
```