36:题目描述
统计一个数字在排序数组中出现的次数。
思路:二分法查找
class Solution {
public:
int process(vector<int>&data,float k)
{
int left=0,right=data.size()-1;
int mid=0;
while(left<=right)
{
mid=(left+right)>>1;
if(data[mid]>k)
right=mid-1;
else if(data[mid]<k)
left=mid+1;
}
return left;
}
int GetNumberOfK(vector<int> data ,int k) {
if(data.empty())
return 0;
int left=process(data,k-0.5);
int right=process(data,k+0.5);
return right-left;
}
};
# -*- coding:utf-8 -*-
class Solution:
def GetNumberOfK(self, data, k):
return data.count(k)
# write code here
37.题目描述
输入一棵二叉树,求该树的深度。从根结点到叶结点依次经过的结点(含根、叶结点)形成树的一条路径,最长路径的长度为树的深度。
思路:递归法,深度优先搜索dfs,从底部开始搜索,每次返回节点的左右子树最大值加一,加一是因为当前节点的深度。
层序法也可以
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
int TreeDepth(TreeNode* pRoot)
{
if(pRoot==NULL)
return 0;
int l_left=TreeDepth(pRoot->left);
int l_right=TreeDepth(pRoot->right);
return max(l_left+1,l_right+1);
}
};
class Solution {
public:
int TreeDepth(TreeNode* pRoot)
{
if(pRoot==NULL)
return 0;
queue<TreeNode*>node;
node.push(pRoot);
int deep=0;
while(!node.empty())
{
int len=node.size();
while(len)
{
TreeNode* temp = node.front();
if(temp->left != NULL)
node.push(temp->left);
if(temp->right != NULL)
node.push(temp->right);
node.pop();
--len;
}
++deep;
}
return deep;
}
};
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def TreeDepth(self, pRoot):
if pRoot==None:
return 0
left=self.TreeDepth(pRoot.left)
right=self.TreeDepth(pRoot.right)
return max(left+1,right+1)
# write code here