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CyC2018 2018-03-27 11:49:15 +08:00
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notes/剑指 offer 题解.md
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@ -2460,15 +2460,15 @@ public double countProbability(int n, int s) {
## 解题思路 ## 解题思路
```java ```java
public boolean isContinuous(int [] numbers) { public boolean isContinuous(int[] nums) {
if(numbers.length < 5) return false; if (nums.length < 5) return false;
Arrays.sort(numbers); Arrays.sort(nums);
int cnt = 0; int cnt = 0;
for(int num : numbers) if(num == 0) cnt++; for (int num : nums) if (num == 0) cnt++;
for(int i = cnt; i < numbers.length - 1; i++) { for (int i = cnt; i < nums.length - 1; i++) {
if(numbers[i + 1] == numbers[i]) return false; if (nums[i + 1] == nums[i]) return false;
int interval = numbers[i + 1] - numbers[i] - 1; int interval = nums[i + 1] - nums[i] - 1;
if(interval > cnt) return false; if (interval > cnt) return false;
cnt -= interval; cnt -= interval;
} }
return true; return true;
@ -2583,7 +2583,18 @@ public int StrToInt(String str) {
# 68. 树中两个节点的最低公共祖先 # 68. 树中两个节点的最低公共祖先
树是二叉查找树的最低公共祖先问题： ## 二叉查找树
```html
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
```
```java ```java
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
@ -2592,3 +2603,25 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
return root; return root;
} }
``` ```
## 普通二叉树
```html
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
```
```java
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q) return root;
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
return left == null ? right : right == null ? left : root;
}
```