每日一题day2：LeetCode
面试题 16.05. 阶乘尾数
面试题 16.15. 珠玑妙算
面试题 10.05. 稀疏数组搜索
面试题 05.01. 插入
19. 删除链表的倒数第 N 个结点
2. 两数相加
34. 在排序数组中查找元素的第一个和最后一个位置
1、面试题 16.05. 阶乘尾数
class Solution {
public:
int trailingZeroes(int n) {
int ans = 0;
while(n >= 5){
ans += n / 5;
n /= 5;
}
return ans;
}
};
2、面试题 16.15. 珠玑妙算
class Solution {
public:
int ans[2];
vector<int> masterMind(string solution, string guess) {
map<char, int> cnt1, cnt2;
ans[0] = ans[1] = 0;
int len1 = solution.size();
for(int i = 0; i < len1; i++){
if(solution[i] == guess[i]) ans[0]++;
else{
cnt1[solution[i]]++;
cnt2[guess[i]]++;
}
}
for(int i = 0; i < len1; i++){
if(cnt1.count(guess[i]) != 0){
ans[1] += min(cnt1[guess[i]], cnt2[guess[i]]);
cnt1[guess[i]] = cnt2[guess[i]] = 0;
}
}
return {
ans[0], ans[1]};
}
};
3、面试题 10.05. 稀疏数组搜索
class Solution {
public:
int findString(vector<string>& words, string s) {
int l = 0, r = words.size() - 1, ans = -1;
while(l <= r){
if(words[l].size() == 0){
l++; continue;
}
if(words[r].size() == 0){
r--; continue;
}
int mid = (l + r) >> 1;
while(words[mid].size() == 0){
mid--;
if(mid == l){
l = (l + r) >> 1;
continue;
}
}
if(words[mid] == s){
ans = mid; break; }
else if(words[mid] > s) r = mid - 1;
else l = mid + 1;
}
return ans;
}
};
4、面试题 05.01. 插入
class Solution {
public:
int insertBits(int N, int M, int i, int j) {
if(!N) return M;
for(int k = i; k <= j; k++){
int temp = M % 2; M /= 2;
if(temp){
N |= (1 << k);
}else{
if(N & (1 << k)) N -= (1 << k);
}
}
return N;
}
};
5、19. 删除链表的倒数第 N 个结点
/**
* 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* removeNthFromEnd(ListNode* head, int n) {
ListNode* p = head;
ListNode* q = head;
ListNode* HEAD = new ListNode(1);
HEAD -> next = head;
ListNode* pre = HEAD;
while(n-- && p) p = p -> next;
if(n != -1) return head;
while(p){
pre = q;
p = p -> next;
q = q -> next;
}
pre -> next = pre -> next -> next;
return HEAD -> next;
}
};
6、2. 两数相加
/**
* 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* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* HEAD = new ListNode(1);
ListNode* p = l1;
ListNode* q = l2;
ListNode* now = HEAD;
int yu = 0, jin = 0;
while(p && q){
yu = (p -> val + q -> val + jin) % 10;
jin = (p -> val + q -> val + jin) / 10;
ListNode* temp = new ListNode(yu);
now -> next = temp; now = now -> next;
p = p -> next; q = q -> next;
}
while(p){
yu = (p -> val + jin) % 10;
jin = (p -> val + jin) / 10;
ListNode* temp = new ListNode(yu);
now -> next = temp; now = now -> next;
p = p -> next;
}
while(q){
yu = (q -> val + jin) % 10;
jin = (q -> val + jin) / 10;
ListNode* temp = new ListNode(yu);
now -> next = temp; now = now -> next;
q = q -> next;
}
if(jin){
ListNode* temp = new ListNode(jin);
now -> next = temp; now = now -> next;
}
return HEAD -> next;
}
};
7、34. 在排序数组中查找元素的第一个和最后一个位置
class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
nums.push_back(INT_MAX);
int pos1 = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
int pos2 = upper_bound(nums.begin(), nums.end(), target) - nums.begin() - 1;
if(nums[pos1] > target) return {
-1, -1};
else return {
pos1, pos2};
}
};
2021/3/6完结。