LeetCode刷算法题-简单难度（九）题号801-900
804.唯一摩斯密码词
class Solution {
public:
int uniqueMorseRepresentations(vector<string>& words) {
vector<string> table{
".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
set<string> tmp;
for(auto s : words){
string trans;
for(auto c : s){
trans += table[c-'a'];
}
tmp.insert(trans);
}
return tmp.size();
}
};
806.写字符串需要的行数
class Solution {
public:
vector<int> numberOfLines(vector<int>& widths, string S) {
int lines = 1, w = 0;
for(auto c : S){
w += widths[c-'a'];
if(w > 100){
lines++;
w = widths[c-'a'];
}
}
vector<int> res{
lines,w};
return res;
}
};
811.子域名访问计数
class Solution {
public:
vector<string> subdomainVisits(vector<string>& cpdomains) {
vector<string> res;
int n = cpdomains.size();
map<string, int> tmp;
int idx0,idx1,idx2;
for(int i = 0; i < n; ++i){
int m = cpdomains[i].size();
int count = 0;
for(int j = 0; j < m; ++j){
if(' ' == cpdomains[i][j]){
idx0 = j;
}
if(0==count){
if('.'==cpdomains[i][j]){
count++;
idx1 = j;
}
}else if(1==count){
if('.'==cpdomains[i][j]){
count++;
idx2 = j;
}
}else break;
}
int times = stoi(cpdomains[i].substr(0,idx0));
tmp[cpdomains[i].substr(idx1+1,m-idx1-1)] += times;
tmp[cpdomains[i].substr(idx0+1,m-idx0-1)] += times;
if(2==count){
tmp[cpdomains[i].substr(idx2+1,m-idx2-1)] += times;
}
}
for(auto iter = tmp.begin(); iter != tmp.end(); ++iter){
string str = to_string(iter->second) + " " + iter->first;
res.push_back(str);
}
return res;
}
};
812.最大三角形面积
class Solution {
public:
double largestTriangleArea(vector<vector<int>>& points) {
double s = 0;
for(int i =0;i<points.size();i++){
for(int j=0;j<points.size();j++){
for(int k=0;k<points.size();k++){
s = max(s,0.5*(points[k][0]*points[j][1]+points[j][0]*points[i][1]+points[i][0]*points[k][1]-points[j][1]*points[i][0]-points[k][1]*points[j][0]-points[k][0]*points[i][1]));
}
}
}
return s;
}
};
819.最常见的词
class Solution {
public:
string mostCommonWord(string paragraph, vector<string>& banned) {
unordered_map<string,int> count;
for(auto& c : paragraph){
c = isalpha(c) ? tolower(c) : ' ';
}
istringstream iss(paragraph);
string w;
pair<string,int> res{
"",0};
while(iss >> w){
if(find(banned.begin(), banned.end(),w)==banned.end() && ++count[w] > res.second){
res = make_pair(w, count[w]);
}
}
return res.first;
}
};
821.字符的最短距离
class Solution {
public:
vector<int> shortestToChar(string S, char C) {
int n = S.length();
vector<int> res(n);
int idx = -1;
for(int i = 0; i < n; ++i){
if(S[i] == C){
for(int k = idx + 1; k < i; ++k){
res[k] = idx == -1 ? (i - k) : min(k-idx, i - k);
}
res[i] = 0;
idx = i;
}
}
if(idx < n - 1){
for(int k = idx + 1; k < n; ++k){
res[k] = k - idx;
}
}
return res;
}
};
824.山羊拉丁文
class Solution {
public:
string toGoatLatin(string S) {
string vo = "aeiouAEIOU";
istringstream iss(S);
string w;
string res;
int count = 0;
while(iss>>w){
count++;
if(vo.find(w[0]) == string::npos){
w.push_back(w[0]);
w.erase(w.begin());
}
res += w + "ma" + string(count,'a') + " ";
}
res.erase(res.find_last_of(' '));
return res;
}
};
830.较大分组的位置
class Solution {
public:
vector<vector<int>> largeGroupPositions(string S) {
char last = S[0];
int beg = 0, end = 0;
vector<vector<int>> res;
for(int i = 1; i < S.size(); ++i){
if(S[i] == last) end++;
else{
if(end-beg>=2) {
res.push_back(vector<int>{
beg,end});
}
beg = end = i;
last = S[i];
}
}
if(end-beg>=2) res.push_back(vector<int>{
beg,end});
return res;
}
};
832.翻转图像
class Solution {
public:
vector<vector<int>> flipAndInvertImage(vector<vector<int>>& A) {
int n = A.size(), m = A[0].size();
for(int i = 0; i < n; ++i)
for(int j = 0; j < (m+1)/2; ++j){
int tmp = A[i][j];
A[i][j] = A[i][m-1-j] ^ 1;
A[i][m-1-j] = tmp ^ 1;
}
return A;
}
};
836.矩形重叠
class Solution {
public:
bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {
int bx = max(rec1[0],rec2[0]);
int tx = min(rec1[2],rec2[2]);
int by = max(rec1[1],rec2[1]);
int ty = min(rec1[3],rec2[3]);
return (bx < tx) && (by < ty);
}
};
840.矩阵中的幻方
class Solution {
public:
int numMagicSquaresInside(vector<vector<int>>& grid) {
if(grid.size() < 3 || grid[0].size() < 3) return 0;
int res = 0;
int n = grid.size(), m = grid[0].size();
for(int i = 0; i < n - 2; ++i)
for(int j = 0; j < m - 2; ++j){
set<int> sgrid;
int k = 0;
for(; k < 3; ++k){
for(int l = 0; l < 3; ++l){
sgrid.insert(grid[i+k][j+l]);
}
if(grid[i+k][j] + grid[i+k][j+1] + grid[i+k][j+2] != 15) break;
}
if(k!=3) continue;
if(sgrid.size() < 9 || *(--sgrid.end()) > 9) continue;
if(grid[i][j] + grid[i+1][j+1] + grid[i+2][j+2] != 15) continue;
if(grid[i][j] + grid[i+1][j] + grid[i+2][j] != 15) continue;
res++;
}
return res;
}
};
844.比较含退格的字符串
class Solution {
public:
bool backspaceCompare(string S, string T) {
stack<char> ss,st;
for(auto c : S){
if('#'==c){
if(ss.empty()) continue;
ss.pop();
}else ss.push(c);
}
for(auto c : T){
if('#'==c){
if(st.empty()) continue;
st.pop();
}else st.push(c);
}
return ss == st;
}
};
849.到最近的人的最大距离
class Solution {
public:
int maxDistToClosest(vector<int>& seats) {
int n = seats.size();
int maxLen = 0;
int i = 0;
while(i<n){
if(seats[i]==0){
int k = 0;
int l = i - 1;
while(l>=0&&seats[l]==0) l--;
int r = i + 1;
while(r<n&&seats[r]==0) r++;
if(l < 0) k = r - i;
else if(r==n) k = i - l;
else k = min(r-i,i-l);
maxLen = max(maxLen,k);
}
++i;
}
return maxLen;
}
};
852.山峰数组的峰顶索引
class Solution {
public:
int peakIndexInMountainArray(vector<int>& A) {
int left = 0, right = A.size()-1, mid;
while(left < right){
mid = (left + right) / 2;
if(A[mid-1] < A[mid] && A[mid] < A[mid+1]) left = mid;
else if(A[mid-1] > A[mid] && A[mid] > A[mid+1]) right = mid;
else break;
}
return mid;
}
};
859.亲密字符串
class Solution {
public:
bool buddyStrings(string A, string B) {
if(A.size() != B.size())
return false;
vector<int> temp;
for(int i=0;i<A.length();i++){
if(A[i] !=B[i]){
temp.push_back(i);
if(temp.size()>2){
return false;
}
}
}
if(temp.size() == 0){
map<char,int> m;
for(int i=0;i<A.length();i++){
m[A[i]]++;
if(m[A[i]]>=2){
return true;
}
}
return false;
}
if(temp.size() == 1)
return false;
else{
if(A[temp[0]] == B[temp[1]] && A[temp[1]]== B[temp[0]]){
return true;
}
return false;
}
}
};
860.柠檬水找零
class Solution {
public:
bool lemonadeChange(vector<int>& bills) {
int fives = 0, tens = 0;
for(auto n : bills){
switch(n){
case 5:
fives++;
break;
case 10:
fives--;
tens++;
break;
case 20:
tens > 0 ? tens--,fives-- : fives -= 3;
break;
}
if(fives < 0) return false;
}
return true;
}
};
867.转置矩阵
class Solution {
public:
vector<vector<int>> transpose(vector<vector<int>>& A) {
int n = A.size(), m = A[0].size();
vector<vector<int>> res(m, vector<int>(n));
for(int i = 0; i < m; ++i){
for(int j = 0; j < n; ++j){
res[i][j] = A[j][i];
}
}
return res;
}
};
868.二进制间距
class Solution {
public:
int binaryGap(int N) {
int idx = -1, count = 0;
int res = 0;
while(N){
if(N&1){
if(idx != -1) res = max(res, count - idx);
idx = count;
}
N >>= 1;
count++;
}
return res;
}
};
872.叶子相似的树
/**
* 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:
bool leafSimilar(TreeNode* root1, TreeNode* root2) {
vector<int> lv1, lv2;
lvSeq(lv1,root1);
lvSeq(lv2,root2);
return lv1==lv2;
}
void lvSeq(vector<int>& lv, TreeNode* root){
if(root==NULL) return;
stack<TreeNode*> s;
s.push(root);
while(!s.empty()){
TreeNode* p = s.top();
s.pop();
if(p->left == NULL && p->right == NULL) lv.push_back(p->val);
if(p->right!=NULL) s.push(p->right);
if(p->left!=NULL) s.push(p->left);
}
return;
}
};
874.模拟行走机器人
/*
*在set中查找的速度要远比vector快
*/
class Solution {
public:
int robotSim(vector<int>& commands, vector<vector<int>>& obstacles) {
int res = 0;
vector<pair<int,int>> dir{
{
0,1},{
1,0},{
0,-1},{
-1,0}};
int i = 0, x = 0, y = 0;
set<pair<int,int>> ob;
for(auto o : obstacles){
ob.insert(make_pair(o[0],o[1]));
}
for(auto cd : commands){
if(-2==cd) i = (i+3)%4;
else if(-1==cd) i = (++i)%4;
else {
while(cd){
int dx = x + dir[i].first;
int dy = y + dir[i].second;
if(ob.find(make_pair(dx,dy))==ob.end()){
x = dx;
y = dy;
res = max(res, x*x+y*y);
}else break;
cd--;
}
}
}
return res;
}
};
876.链表的中间节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* middleNode(ListNode* head) {
ListNode * fast = head;
while(!(fast==NULL || fast->next==NULL)){
head = head->next;
fast = fast->next->next;
}
return head;
}
};
883.三维形体投影面积
class Solution {
public:
int projectionArea(vector<vector<int>>& grid) {
int sum = 0;
for(int i = 0; i < grid.size(); ++i){
int temp1 = 0, temp2 = 0;
for(int j = 0; j < grid[0].size(); ++j){
if(grid[i][j]>0) sum++;
temp1 = max(temp1, grid[i][j]);
temp2 = max(temp2, grid[j][i]);
}
sum += temp1 + temp2;
}
return sum;
}
};
884.两句话中的不常见单词
class Solution {
public:
vector<string> uncommonFromSentences(string A, string B) {
string t = A + " " + B;
istringstream iss(t);
string w;
map<string, int> mmap;
vector<string> res;
while(iss >> w){
mmap[w]++;
}
for(auto iter = mmap.begin(); iter != mmap.end(); ++iter){
if(1==iter->second)
res.push_back(iter->first);
}
return res;
}
};
888.公平的糖果交换
class Solution {
public:
vector<int> fairCandySwap(vector<int>& A, vector<int>& B) {
int m = A.size(), n = B.size();
vector<int> res(2);
int sum1 = 0, sum2 = 0;
int l = max(m, n);
for(int i = 0; i < l; ++i){
if(i < m) sum1 += A[i];
if(i < n) sum2 += B[i];
}
unordered_map<int,int> amap;
for(auto a : A){
amap[(sum2 - sum1)/2 + a] = 1;
}
for(auto b : B){
if(amap.find(b)!=amap.end()){
res[0] = (sum1 - sum2)/2 + b;
res[1] = b;
return res;
}
}
return res;
}
};
892.三维形体的表面积
class Solution {
public:
int surfaceArea(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
int area = 0;
for(int i = 0; i < n; ++i){
for(int j = 0; j < m; ++j){
if(grid[i][j] > 0) area += 4*grid[i][j]+2;
if(i < n - 1) area -= min(grid[i][j], grid[i+1][j])*2;
if(j < m - 1) area -= min(grid[i][j], grid[i][j+1])*2;
}
}
return area;
}
};
893.特殊等价字符串
class Solution {
public:
int numSpecialEquivGroups(vector<string>& A) {
set<string> m;
for(int i = 0; i < A.size(); ++i){
string odd = "";
string even = "";
for(int j = 0; j < A[i].length(); ++j){
if(j%2==0) odd += A[i][j];
else even += A[i][j];
}
sort(odd.begin(), odd.end());
sort(even.begin(), even.end());
m.insert(odd+even);
}
return m.size();
}
};
896.单调数列
class Solution {
public:
bool isMonotonic(vector<int>& A) {
bool inc = true, dec = true;
int n = A.size();
for(int i = 1; i < n; ++i){
if(inc || dec){
if(A[i] > A[i-1]) dec = false;
if(A[i] < A[i-1]) inc = false;
} else break;
}
return inc || dec;
}
};
897.递增顺序查找树
/**
* 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:
TreeNode* increasingBST(TreeNode* root) {
queue<TreeNode*> qt;
inOrder(root, qt);
TreeNode* head = qt.front();
qt.pop();
TreeNode* p = head;
while(!qt.empty()){
p->left = NULL;
p->right = qt.front();
qt.pop();
p = p->right;
}
p->left = p->right = NULL;
return head;
}
void inOrder(TreeNode* root, queue<TreeNode*>& qt){
if(root != NULL){
inOrder(root->left, qt);
qt.push(root);
inOrder(root->right,qt);
}
}
};