LeetCode Weekly Contest 19
LeetCode Weekly Contest 19
504. Base 7
- User Accepted: 994
- User Tried: 1054
- Total Accepted: 1345
- Total Submissions: 3160
- Difficulty: Easy
Given an integer, return its base 7 string representation.
Example 1:
Input: 100 Output: "202"
Example 2:
Input: -7 Output: "-10"
Note: The input will be in range of [-1e7, 1e7].
简单的十进制转7进制。
class Solution {
public:
string convertToBase7(int num) {
if(num == 0){
return "0";
}
string s, p="";
if(num < 0){
p = "-";
num = -num;
}
while(num){
s = char('0' + (num%7)) + s;
num = num/7;
}
return p+s;
}
};
513. Find Bottom Left Tree Value
- User Accepted: 870
- User Tried: 943
- Total Accepted: 1060
- Total Submissions: 2399
- Difficulty: Medium
Given a binary tree, find the leftmost value in the last row of the tree.
Example 1:
Input:
2
/ \
1 3
Output:
1
Example 2:
Input:
1
/ \
2 3
/ / \
4 5 6
/
7
Output:
7
Note: You may assume the tree (i.e., the given root node) is not NULL.
树结构的迭代
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int ans, current;
void Find(TreeNode *root, int row){
if(row >= current){
ans = root->val;
current = row;
}
if(root->right != NULL){
Find(root->right, row+1);
}
if(root->left != NULL){
Find(root->left, row+1);
}
}
int findBottomLeftValue(TreeNode* root) {
ans = root->val; current = 1;
Find(root, 1);
return ans;
}
};
515. Find Largest Value in Each Tree Row
- User Accepted: 851
- User Tried: 878
- Total Accepted: 943
- Total Submissions: 2140
- Difficulty: Medium
You need to find the largest value in each row of a binary tree.
Example:
Input:
1
/ \
3 2
/ \ \
5 3 9
Output: [1, 3, 9]
也是用树结构，使用 BFS
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> largestValues(TreeNode* root) {
vector<int> ans;
if(root == NULL){
return ans;
}
TreeNode *tmp;
queue<TreeNode* > q;
queue<int> qlevel;
qlevel.push(1);
q.push(root);
int cur_level = 1, cur_val = -2147483648;
while(!q.empty()){
tmp = q.front(); q.pop();
int tmp_level = qlevel.front(); qlevel.pop();
if(tmp_level == cur_level){
if(cur_val < tmp->val){
cur_val = tmp->val;
}
}else{
ans.push_back( cur_val );
cur_level = tmp_level;
cur_val = tmp->val;
}
if(tmp->left != NULL){
q.push(tmp->left); qlevel.push(tmp_level + 1);
}
if(tmp->right != NULL){
q.push(tmp->right); qlevel.push(tmp_level + 1);
}
}
ans.push_back(cur_val);
return ans;
}
};
493. Reverse Pairs
- User Accepted: 165
- User Tried: 777
- Total Accepted: 172
- Total Submissions: 3595
- Difficulty: Hard
Given an array nums, we call (i, j) an important reverse pair if i < j and nums[i] > 2*nums[j].
You need to return the number of important reverse pairs in the given array.
Example1:
Input: [1,3,2,3,1] Output: 2
Example2:
Input: [2,4,3,5,1] Output: 3
Note:
- The length of the given array will not exceed
50,000. - All the numbers in the input array are in the range of 32-bit integer.
看了discuss 才有思路， 原来 merge sort 也有这么好功能
class Solution {
private:
int cnt;
void countPair(vector<int>& nums, int lf, int md, int rg){
int l = lf, r = md + 1;
while(l <= md && r <= rg){
if((long)nums[l] > (long)2*nums[r]){
cnt += (md - l + 1);
r++;
}else{
l++;
}
}
sort(nums.begin() + lf, nums.begin() + rg + 1);
}
void Reverse(vector<int>& nums, int lf, int rg){
if(lf == rg){
return;
}
int mid = lf + (rg - lf)/2;
Reverse(nums, lf, mid);
Reverse(nums, mid+1, rg);
countPair(nums, lf, mid, rg);
}
public:
int reversePairs(vector<int>& nums) {
int len = nums.size();
if(len <= 1){
return 0;
}
cnt = 0;
Reverse(nums, 0, len-1);
return cnt;
}
};
