LeetCode题解-04（位运算、排序、数学）
目录
- LeetCode题解
- chap-10:位运算
- chap-11:排序
- chap-12:数学
- 1、字符串相乘
- 2、Pow(x, n)【快速幂】
- 3、排列序列【康托展开】
- 4、二进制求和
- 5、格雷编码
- 6、直线上最多的点数
- 7、分数到小数【循环节】
- 8、阶乘后的零
- 9、最大数
- 10、快乐数【快慢指针】
- 11、数字 1 的个数
- 12、丑数
- 13、计数质数【筛质数】
- 14、矩形面积
- 15、各位相加
- 16、Nim 游戏【博弈论】
- 17、灯泡开关
- 18、2 的幂
- 19、3 的幂
- 20、4的幂
- 21、整数拆分
- 22、计算各个位数不同的数字个数
- 23、水壶问题
- 24、整数替换
- 25、Fizz Buzz
- 26、等差数列划分
- 27、从英文中重建数字
- 28、字典序的第K小数字
- 29、排列硬币
- 30、重复的子字符串【kmp】
- 31、最少移动次数使数组元素相等 II
- 32、岛屿的周长
- 33、构造矩形
- 34、n 的第 k 个因子
- 35、使数组中所有元素相等的最小操作数
- 36、第 K 条最小指令【组合排列迭代+字典树】
LeetCode题解
chap-10:位运算
1、二进制求和
class Solution {
public:
string addBinary(string a, string b) {
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
string ans;
for(int i=0,j=0,t=0;t!=0 || i<a.size() || j<b.size();){
if(i<a.size()) t+=a[i++]-'0';
if(j<b.size()) t+=b[j++]-'0';
ans += ((t%2) + '0');
t/=2;
}
reverse(ans.begin(), ans.end());
return ans;
}
};
2、只出现一次的数字
class Solution {
public:
int singleNumber(vector<int>& nums) {
int ans = 0;
for(auto &c:nums) ans^=c;
return ans;
}
};
3、只出现一次的数字 II
class Solution {
public:
int singleNumber(vector<int>& nums) {
int ans = 0;
for(int i=0;i<32;i++){
int sum=0;
for(int j=0;j<nums.size();j++){
sum+=(nums[j]>>i)&1;
}
ans += ((sum%3)<<i);
}
return ans;
}
};
4、只出现一次的数字 III【分组异或】
class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
int t = nums[0], i=0;
for(i=1;i<nums.size();i++) t^=nums[i];
for(i=0;i<32;i++)
if((t>>i)&1 == 1) break;
int a=0,b=0;
for(auto &c:nums){
if((c>>i)&1) a^=c;
else b^=c;
}
return vector<int>({a,b});
}
};
5、丢失的数字
class Solution {
public:
int missingNumber(vector<int>& nums) {
int ans=0;
for(int i=0;i<=nums.size();i++) ans^=i;
for(auto &c:nums) ans^=c;
return ans;
}
};
6、生命游戏
如何做到不用额外的空间，且把所有位置细胞的状态同时改变呢？因为想到只有0和1两个状态，可以用二进制中的第二位来存储变化后的状态。
0000：一开始是死细胞，后来还是死细胞
0101：一开始是活细胞，后来变成死细胞
1010：一开始是死细胞，后来变成活细胞
1111：一开始是活细胞，后来还是活细胞
class Solution {
public:
void gameOfLife(vector<vector<int>>& board) {
if (board.empty() || board[0].empty()) return;
int n = board.size(), m = board[0].size();
for (int i = 0; i < n; i ++ )
for (int j = 0; j < m; j ++ ) {
int live = 0;
for (int x = max(0, i - 1); x <= min(n - 1, i + 1); x ++ )
for (int y = max(0, j - 1); y <= min(m - 1, j + 1); y ++ )
if ((x != i || y != j) && (board[x][y] & 1))
live ++ ;
int cur = board[i][j] & 1, next;
if (cur) {
if (live < 2 || live > 3) next = 0;
else next = 1;
} else {
if (live == 3) next = 1;
else next = 0;
}
board[i][j] |= next << 1;
}
for (int i = 0; i < n; i ++ )
for (int j = 0; j < m; j ++ )
board[i][j] >>= 1;
}
};
7、最大单词长度乘积
class Solution {
public:
int maxProduct(vector<string>& words) {
vector<int> state;
for (auto word: words) {
int s = 0;
for (auto c: word)
s |= 1 << (c - 'a');
state.push_back(s);
}
int res = 0;
for (int i = 0; i < words.size(); i ++ )
for (int j = i + 1; j < words.size(); j ++ )
if ((state[i] & state[j]) == 0)
res = max(res, (int)(words[i].size() * words[j].size()));
return res;
}
};
8、两整数之和
class Solution {
public:
int getSum(int a, int b) {
if(a == 0) return b;
if(b == 0) return a;
int sum = a ^ b;
int carry = unsigned(a & b) << 1;
return getSum(sum, carry);
}
};
9、找不同
class Solution {
public:
char findTheDifference(string s, string t) {
char ans=0;
for(auto c:s) ans^=c;
for(auto c:t) ans^=c;
return ans;
}
};
10、数字转换为十六进制数
class Solution {
public:
string toHex(unsigned int n) {
if(!n) return "0";
string ans, nums = "0123456789abcdef";
while(n){
ans = nums[n&0xf] + ans;
n=n>>4;
}
return ans;
}
};
11、汉明距离
class Solution {
public:
int hammingDistance(int x, int y) {
bitset<32> t(x^y);int ans = 0;
for(int i=0;i<32;i++) if(t[i] == 1) ans++;
return ans;
}
};
12、数字的补数
class Solution {
public:
int findComplement(int num) {
int cnt = 0;
for (int x = num; x; x >>= 1) cnt ++ ;
return ~num & ((1ll << cnt) - 1); // 消除高位
}
};
13、汉明距离总和
class Solution {
public:
int totalHammingDistance(vector<int>& nums) {
int n=nums.size(), ans=0;
for(int i=0;i<32;i++){
int ones = 0;
for(auto t:nums){
if((1ll<<i)&t) ones++;
}
ans+=ones*(n-ones);
}
return ans;
}
};
14、连接连续二进制数字
class Solution {
public:
int find(int n){
for(int i=0;i<31;i++){
if(n >= (1<<i)) continue;
else return i;
}
return -1;
}
int concatenatedBinary(int n) {
const static int N=1e9+7;
long long ans = 0;
for(int i=1;i<=n;i++){
ans = ans%N;
int high=find(i);
ans<<=high;
ans+=i;
}
int t = ans%N;
return t;
}
};
[Go Back~~](# LeetCode题解)
chap-11:排序
1、对链表进行插入排序
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* insertionSortList(ListNode* head) {
auto dummy = new ListNode(INT_MIN);
dummy->next = head;
auto cur = dummy;
while(cur->next){
if(cur->next->val >= cur->val){
cur = cur->next;
}else{
auto t = cur->next;
cur->next = t->next;
auto pre = dummy;
while(pre->next->val <= t->val) pre = pre->next;
t->next = pre->next;
pre->next = t;
}
}
return dummy->next;
}
};
2、排序链表【×】
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* merge(ListNode* l, ListNode* r){
auto dummy = new ListNode(0);
auto t = dummy;
while(l && r){
if(l->val <= r->val){
t->next = l; l = l->next; t = t->next;
}else{
t->next = r; r = r->next; t = t->next;
}
}
while(l) {t->next = l; l = l->next; t = t->next;}
while(r) {t->next = r; r = r->next; t = t->next;}
t = dummy->next;
delete dummy;
return t;
}
ListNode* sortList(ListNode* head) {
if(head == nullptr || head->next == nullptr) return head; //
auto slow = head; auto fast = head;
while(fast->next && fast->next->next){ //
slow = slow->next;
fast = fast->next->next;
}
fast = slow->next;
slow->next = nullptr;
auto l = sortList(head);
auto r = sortList(fast);
return merge(l,r);
}
};
3、丑数 II【多路归并】
class Solution {
public:
int nthUglyNumber(int n) {
auto help = new int[n];
help[0] = 1;
int f2,f3,f5;
f2 = f3 = f5 = 0;
for(int i=1;i<n;i++){
help[i] = min(min(help[f2]*2,help[f3]*3),help[f5]*5);
if(help[f2]*2 == help[i]) f2++;
if(help[f3]*3 == help[i]) f3++;
if(help[f5]*5 == help[i]) f5++;
}
return help[n-1];
}
};
4、超级丑数
class Solution {
public:
int nthSuperUglyNumber(int n, vector<int>& primes) {
int m = primes.size();
vector<int> f(m,0);
auto help = new int[n];
help[0] = 1;
for(int i=1;i<n;i++){
int t = INT_MAX;
for(int j=0;j<m;j++) t = min(help[f[j]]*primes[j],t);
help[i] = t;
for(int j=0;j<m;j++){
if(t == help[f[j]]*primes[j]) f[j]++;
}
}
return help[n-1];
}
};
5、排序数组【模板题】
// 归并排序
class Solution {
public:
void merge(vector<int>&nums,int l1,int l2,int r1,int r2){
vector<int> temp(r2-l1+1,0);
int idx = 0, a = l1;
while(l1<=l2 && r1<=r2){
if(nums[l1] < nums[r1]) temp[idx++] = nums[l1++];
else temp[idx++] = nums[r1++];
}
while(l1<=l2) temp[idx++] = nums[l1++];
while(r1<=r2) temp[idx++] = nums[r1++];
idx = 0;
while(idx<temp.size()) nums[a++] = temp[idx++];
}
void sort_(vector<int>& nums,int l,int r){
if(l>=r) return;
int mid = l+r>>1;
sort_(nums,l,mid); sort_(nums,mid+1,r);
merge(nums,l,mid,mid+1,r);
}
vector<int> sortArray(vector<int>& nums) {
int n = nums.size();
if(n > 1) sort_(nums,0,n-1);
return nums;
}
};
// 快速排序
class Solution {
public:
void sort_(vector<int>&nums,int l,int r){
if(l>=r) return;
int i=l-1, j=r+1, x = nums[l+r>>1];
while(i<j){
do i++; while(nums[i]<x);
do j--; while(nums[j]>x);
if(i<j) swap(nums[i],nums[j]);
}
sort_(nums,l,j);
sort_(nums,j+1,r);
}
vector<int> sortArray(vector<int>& nums) {
int n = nums.size();
if(n > 1)
sort_(nums,0,n-1);
return nums;
}
};
6、数组的相对排序
class Solution {
public:
vector<int> relativeSortArray(vector<int>& a, vector<int>& b) {
unordered_map<int,int> hash;
int n = b.size();
for(int i=0;i<n;i++) hash[b[i]]=i;
auto cmp = [&](int x,int y){
int t1 = n+x;
int t2 = n+y;
if(hash.count(x)) t1 = hash[x];
if(hash.count(y)) t2 = hash[y];
return t1<t2;
};
sort(a.begin(),a.end(),cmp);
return a;
}
};
7、数组中两元素的最大乘积
class Solution {
public:
int maxProduct(vector<int>& nums) {
int a = INT_MIN, b = INT_MIN;
for(auto t:nums){
if(t > a){
b = a;
a = t;
}else if(t>b) b = t;
}
return (a-1)*(b-1);
}
};
8、切割后面积最大的蛋糕
class Solution {
public:
int maxArea(int h, int w, vector<int>& hc, vector<int>& vc) {
const static int MOD=1e9+7;
sort(hc.begin(), hc.end()); hc.push_back(h);
sort(vc.begin(), vc.end()); vc.push_back(w);
int ht = hc[0], vt = vc[0];
for(int i=1;i<hc.size();i++) ht = max(ht, hc[i]-hc[i-1]);
for(int i=1;i<vc.size();i++) vt = max(vt, vc[i]-vc[i-1]);
return (long long)(ht)*vt%MOD;
}
};
9、按照频率将数组升序排序
class Solution {
public:
vector<int> frequencySort(vector<int>& nums) {
unordered_map<int,int> hash;
for(int t:nums) hash[t]++;
auto cmp = [&](int x,int y){
if(hash[x] < hash[y]) return true;
else if(hash[x] == hash[y]){
return x>y;
}else return false;
};
sort(nums.begin(),nums.end(),cmp);
return nums;
}
};
10、两点之间不包含任何点的最宽垂直面积
class Solution {
public:
int maxWidthOfVerticalArea(vector<vector<int>>& points) {
sort(points.begin(), points.end());
const int n = points.size();
int ans = 0;
for (int i = 1; i < n; i++)
if (ans < points[i][0] - points[i - 1][0])
ans = points[i][0] - points[i - 1][0];
return ans;
}
};
[Go Back~~](# LeetCode题解)
chap-12:数学
1、字符串相乘
class Solution {
public:
string multiply(string num1, string num2) {
vector<int>A,B;
int n = num1.size(), m = num2.size();
for(int i = n-1;i>=0;i--) A.push_back(num1[i] - '0');
for(int i = m-1;i>=0;i--) B.push_back(num2[i] - '0');
vector<int> C(m + n);
for(int i = 0;i<n;i++){
for(int j = 0;j<m;j++)
C[i+j] += A[i] * B[j];
}
for(int i = 0,t = 0;i<m+n;i++){
t += C[i];
C[i] = t % 10;
t /= 10;
}
int k = m+n-1;
while(!C[k] && k)k--;
string res;
while(k>=0) res += (C[k--] + '0');
return res;
}
};
2、Pow(x, n)【快速幂】
class Solution {
public:
double myPow(double x, int m) {
long long n = m;
double res = 1;
bool f = false;
if(n < 0) {f = true; n = -n;}
while(n){
if(n & 1) res *= x;
x *= x;
n = n >> 1;
}
return f == true ? 1 / res : res;
}
};
3、排列序列【康托展开】
class Solution {
public:
string getPermutation(int n, int k) {
string ans;
vector<bool> f(n,0);
for(int i=0;i<n;i++){
int t = 1;
for(int j=1;j<n-i;j++) t*=j;
for(int j=0;j<n;j++){
if(!f[j]){
if(t>=k){
f[j] = true;
ans += to_string(j+1);
break;
} k-=t;
}
}
}
return ans;
}
};
4、二进制求和
class Solution {
public:
string addBinary(string a, string b) {
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
string ans;
for(int i=0,j=0,t=0;t!=0 || i<a.size() || j<b.size();){
if(i<a.size()) t+=a[i++]-'0';
if(j<b.size()) t+=b[j++]-'0';
ans += ((t%2) + '0');
t/=2;
}
reverse(ans.begin(), ans.end());
return ans;
}
};
5、格雷编码
class Solution {
public:
// 0 0 0 镜面
// 0 1 0
// 0 1 1
// 0 0 1
vector<int> grayCode(int n) {
vector<int> res = vector<int>(1,0);
while(n--){
for(int i=res.size()-1;i>=0;i--){
res[i] *= 2;
res.push_back(res[i] + 1);
}
}
return res;
}
};
class Solution {
public:
vector<int> grayCode(int n) {
// binary2gray
vector<int> ans;
for(int i=0;i<pow(2,n);i++){
ans.push_back(i^(i>>1));
}
return ans;
}
};
6、直线上最多的点数
class Solution {
public:
int maxPoints(vector<vector<int>>& points) {
typedef long double LD;
int res = 0;
for(int i=0;i<points.size();i++){
int x1 = points[i][0], y1 = points[i][1];
int vs = 0, ss = 0;
unordered_map<LD,int> hash;
for(int j=0;j<points.size();j++){
int x2 = points[j][0], y2 = points[j][1];
if(x2 == x1 && y2 == y1) ss++;
else if(x2 == x1) vs++;
else {
LD k = ((LD)y2-y1)/(x2-x1);
hash[k]++;
}
}
for(auto [k,t]:hash) vs = max(vs,t);
res = max(res, ss + vs);
}
return res;
}
};
7、分数到小数【循环节】
class Solution {
public:
string fractionToDecimal(int numerator, int denominator) {
typedef long long ll;
ll x = numerator,y = denominator;
if(x % y == 0) return to_string(x/y);
string res;
if((x < 0) ^ (y<0)) res += '-';
x = abs(x);y = abs(y);
res += to_string(x / y) + '.'; x %= y;
unordered_map<ll,int>hash;
while(x){
hash[x] = res.size();
x *= 10;
res += to_string(x/y);
x %= y;
if (hash.count(x)) {
res = res.substr(0, hash[x]) + '(' + res.substr(hash[x]) + ')';
break;
}
}
return res;
}
};
8、阶乘后的零
class Solution {
public:
int trailingZeroes(int n) {
int res = 0;
while(n) res += n / 5, n/=5;
return res;
}
};
9、最大数
class Solution {
public:
string largestNumber(vector<int>& nums) {
sort(nums.begin(),nums.end(),[](int a,int b)
{string x = to_string(a),y = to_string(b);return x+y > y+x;});
string res;
for(auto&num:nums) res += to_string(num);
int k = 0;
while(k+1 < res.size() && res[k] == '0')k++;
return res.substr(k);
}
};
10、快乐数【快慢指针】
class Solution {
public:
int get(int x){
int t = 0;
while(x) t+=(x%10)*(x%10), x/=10;
return t;
}
bool isHappy(int n) {
int fast = get(n);
int slow = n;
while(slow != fast){
fast = get(get(fast));
slow = get(slow);
}
return fast == 1;
}
};
11、数字 1 的个数
class Solution {
public:
int countDigitOne(int n) {
if (n <= 0) return 0;
vector<int> nums;
while (n) nums.push_back(n % 10), n /= 10;
reverse(nums.begin(), nums.end());
int res = 0;
for (int i = 0; i < nums.size(); i ++ ) {
int d = nums[i];
int left = 0, right = 0, p = 1;
for (int j = 0; j < i; j ++ ) left = left * 10 + nums[j];
for (int j = i + 1; j < nums.size(); j ++ ) {
right = right * 10 + nums[j];
p = p * 10;
}
if (d == 0) res += left * p;
else if (d == 1) res += left * p + right + 1;
else res += (left + 1) * p;
}
return res;
}
};
12、丑数
class Solution {
public:
bool isUgly(int num) {
if (num <= 0)
return false;
while (num % 2 == 0)
num /= 2;
while (num % 3 == 0)
num /= 3;
while (num % 5 == 0)
num /= 5;
return num == 1;
}
};
13、计数质数【筛质数】
class Solution {
public:
int countPrimes(int n) {
vector<int> primes;
vector<bool> f(n+1,false);
for(int i=2;i<n;i++){
if(f[i] == false) primes.push_back(i);
for(int j=0;i*primes[j] < n;j++){
f[primes[j]*i] = 1;
if(i%primes[j] == 0) break;
}
}
return primes.size();
}
};
14、矩形面积
class Solution {
public:
int computeArea(int ax1, int ay1, int ax2, int ay2, int bx1, int by1, int bx2, int by2) {
int x1 = ax1, y1 = ay2; int x2 = ax2, y2 = ay1;
int x1_ = bx1, y1_ = by2; int x2_ = bx2, y2_ = by1;
ax1 = max(x1,x1_); ay1 = min(y1,y1_);
ax2 = min(x2,x2_); ay2 = max(y2,y2_);
int ans = abs((y2 - y1) * (x2 - x1)) + abs((y2_ - y1_) * (x2_ - x1_));
if(ax1 < ax2 && ay1 > ay2)
return ans - abs((ax2 - ax1)*(ay2 - ay1));
else return ans;
}
};
15、各位相加
class Solution {
public:
int addDigits(int num) {
if(num < 10) return num;
int t = 0;
while(num) t+=num%10, num/=10;
return addDigits(t);
}
};
// O(1)
class Solution {
public:
int addDigits(int num) {
if(!num) return num;
return num%9 ? num%9 : 9;
}
};
16、Nim 游戏【博弈论】
class Solution {
public:
bool canWinNim(int n) {
return n&3; // n%4 != 0
}
};
17、灯泡开关
class Solution {
public:
int bulbSwitch(int n) {
// 约数个数为奇数个则亮
// 约数为奇数表明该数为完全平方数
return sqrt(n);
}
};
18、2 的幂
class Solution {
public:
bool isPowerOfTwo(int n) {
// lowbit
return n>0 ? (n&(-n)) == n : false;
}
};
19、3 的幂
class Solution {
public:
bool isPowerOfThree(int n) {
return n > 0 && 1162261467%n == 0;
}
};
class Solution {
public:
bool isPowerOfThree(int n) {
if(n<1) return false;
while(n){
if(n == 1) return true;
if(n%3) return false;
n/=3;
}
return true;
}
};
20、4的幂
class Solution {
public:
bool isPowerOfFour(int num) {
// 法1:4的整数次幂，等价于 n 是平方数，且 n 的质因子只有2
// return (n>0) && sqrt(n)*sqrt(n) == n && (!((1<<30)%n));
// // 法2:
if (num < 0 || num & (num-1)){//check(is or not) a power of 2.
return false;
}
return num & 0x55555555;//check 1 on odd bits
}
};
21、整数拆分
class Solution {
public:
int integerBreak(int n) {
if(n < 4) return n-1;
int ans = 1;
if(n % 3 == 1) ans*=4, n-=4;
else if(n%3 == 2) ans*=2, n-=2;
while(n) ans*=3, n-=3;
return ans;
}
};
22、计算各个位数不同的数字个数
class Solution {
public:
int countNumbersWithUniqueDigits(int n) {
if(!n) return 1;
n = min(n,10);
vector<int> f(n);
f[0] = 9;
for(int i=1;i<n;i++) f[i] = f[i-1]*(10-i);
int res=0;
for(auto t:f) res+=t; // n中的每个
return res+=1; // 0单独计算
}
};
23、水壶问题
class Solution {
public:
int gcd(int a,int b){
if(b == 0) return a;
return gcd(b, a%b);
}
bool canMeasureWater(int a, int b, int t) {
// 裴蜀定理
if (t > a + b) return false;
a = gcd(a,b);
if(t && t%a == 0) return true;
else return false;
}
};
24、整数替换
class Solution {
public:
int integerReplacement(long long n) {
int ans = 0;
while(n != 1){
ans++;
if((n&3) == 3 && n != 3) n++;
else if(n&1) n--;
else n>>=1;
}
return ans;
}
};
typedef long long LL;
class Solution {
public:
unordered_map<LL, int> dp;
int integerReplacement(int n) {
return f(n);
}
int f(LL n) {
if (dp.count(n)) return dp[n];
if (n == 1) return 0;
if (n % 2 == 0) return dp[n] = f(n / 2) + 1;
return dp[n] = min(f(n + 1), f(n - 1)) + 1;
}
};
25、Fizz Buzz
class Solution {
public:
vector<string> fizzBuzz(int n) {
vector<string> ans;
for(int i=1;i<=n;i++){
string t = "";
if((i%3) == 0) t+="Fizz";
if((i%5) == 0) t+="Buzz";
if(t.empty()) t+=to_string(i);
ans.emplace_back(t);
}
return ans;
}
};
26、等差数列划分
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& nums) {
for (int i = nums.size() - 1; i > 0; i -- ) nums[i] -= nums[i - 1];
int res = 0;
for(int i=1;i<nums.size();i++){
int j=i;
while(j<nums.size() && nums[j] == nums[i])j++;
int len = j-i;
res += len*(len-1)/2;
i=j-1;
}
return res;
}
};
27、从英文中重建数字
class Solution {
public:
string originalDigits(string s) {
string help[] = {"zero","one","two","three","four","five","six","seven","eight","nine"};
int idx[] = {0,2,4,3,5,6,1,7,8,9};
string ans;
unordered_map<char,int> hash;
for(auto c:s) hash[c]++;
for(int i=0;i<10;i++){
int ix = idx[i];
while(true){
bool f = false;
for(int j=0;j<help[ix].size();j++){
if(hash[help[ix][j]] == 0){
while(--j >= 0) hash[help[ix][j]]++;
f = true;
break;
}else {
hash[help[ix][j]]--;
if(j == help[ix].size()-1) ans+=to_string(ix);
}
}
if(f) break;
}
}
sort(ans.begin(),ans.end());
return ans;
}
};
28、字典序的第K小数字
class Solution {
public:
int cal(int pre,int n){
int tot = 0;
long long t = pre, k=1;
while(10*t <= n){
tot+=k;
k*=10, t*=10;
}
if(t <= n){
if(n - t < k) tot+=n-t+1;
else tot+=k;
}
return tot;
}
int findKthNumber(int n, int k) {
int prefix = 1;
while(k>1){
int sz = cal(prefix,n);
if(sz >= k){
prefix*=10;
k--;
}else prefix++, k-=sz;
}
return prefix;
}
};
29、排列硬币
typedef long long ll;
class Solution {
public:
int arrangeCoins(ll n) {
// (1+k)*k / 2 = n -> k^2 + k - 2n = 0
n = n << 3;
int ans = (sqrt(n + 1) - 1) / 2;
return ans;
}
};
30、重复的子字符串【kmp】
class Solution {
public:
bool repeatedSubstringPattern(string s) {
int n = s.size();
vector<int> next(n+1);
s = ' ' + s;
for(int i=2,j=0;i<=n;i++){
while(j && s[i] != s[j+1]) j = next[j];
if(s[i] == s[j+1]) j++;
next[i] = j;
}
int last = n - next[n];
return last<n && n%last == 0;
}
};
31、最少移动次数使数组元素相等 II
class Solution {
public:
int minMoves2(vector<int>& nums) {
int n = nums.size();
nth_element(nums.begin(), nums.begin() + n/2, nums.end());
int t = nums[n/2], ans=0;
for(auto &c:nums) ans+=abs(c-t);
return ans;
}
};
32、岛屿的周长
class Solution {
public:
int islandPerimeter(vector<vector<int>>& grid) {
int ans = 0;
int row[] = {-1,0,1,0};
int col[] = {0,-1,0,1};
for(int i=0;i<grid.size();i++){
for(int j=0;j<grid[0].size();j++){
if(grid[i][j] == 0) continue;
for(int k=0;k<4;k++){
int r = i+row[k], c = j+col[k];
if(r>=0 && r<grid.size() && c >=0 && c<grid[0].size()){
if(grid[r][c] == 0) ans++;
}else ans++;
}
}
}
return ans;
}
};
33、构造矩形
class Solution {
public:
vector<int> constructRectangle(int area) {
int t = sqrt(area);
while(area%t != 0) t--;
return {area/t,t};
}
};
34、n 的第 k 个因子
class Solution {
public:
int kthFactor(int n, int k) {
vector<int> f1,f2;
for(int i=1;i*i<=n;i++){
if(n%i == 0){
f1.push_back(i);
if(i*i != n) f2.push_back(n/i);
}
}
if(k <= f1.size()) return f1[k-1];
if(k <= (f1.size() + f2.size())) return f2[f1.size() + f2.size() - k];
return -1;
}
};
35、使数组中所有元素相等的最小操作数
class Solution {
public:
int minOperations(int n) {
int tot = 0, t = n/2;
while(t--){
tot+= 2*(--n) - 2*t;
}
return tot>>1;
}
};
36、第 K 条最小指令【组合排列迭代+字典树】
class Solution {
public:
string kthSmallestPath(vector<int>& destination, int k) {
int zero = destination[1], one = destination[0];
const int n = one + zero;
vector<vector<int>> C(n+1,vector<int>(n+1));
// 计算排列数
C[0][0] = 1;
for(int i=1;i<=n;i++){
C[i][0] = 1;
for(int j=1;j<=i;j++)
C[i][j] = C[i-1][j] + C[i-1][j-1];
}
string ans;
for(int i=n;i>=1;i--){
if(zero == 0) ans+='V';
else{
if(k > C[i-1][zero-1]){
k-=C[i-1][zero-1];
one--;
ans+='V';
}else{
ans+='H';
zero--;
}
}
}
return ans;
}
};
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