文章目錄
31.整數中1出現的次數
class Solution {
public:
int NumberOf1Between1AndN_Solution(int n)
{
int ret = 0;
for(int i=1; i<=n; i++)
ret += getOneCounts(i);
return ret;
}
private:
int getOneCounts(int n)
{
int ret = 0;
while(n > 0)
{
ret += (n%10==1)?1:0;
n /= 10;
}
return ret;
}
};
32.把數組排成最小的數
對數組進行排序
排序規則：
- 將int a, int b轉爲string
- 若 a＋b<b+a a排在在前
class Solution {
public:
string PrintMinNumber(vector<int> numbers) {
sort(numbers.begin(), numbers.end(), cmp);
string ret;
for(int i=0; i<numbers.size(); i++)
{
ret += to_string(numbers[i]);
}
return ret;
}
private:
static bool cmp(int a, int b)
{
string n1 = to_string(a);
n1 += to_string(b);
string n2 = to_string(b);
n2 += to_string(a);
return n1<n2;
}
};
33. 醜數
暴力法
class Solution {
public:
int GetUglyNumber_Solution(int index) {
if(index == 0) //不加這個報錯！
return 0;
if(index == 1)
return 1;
v = vector<int>(index, 0x3f3f3f3f);
v[0] = 1;
for(int i=1; i<index; i++) //計算第 i+1 個醜數
for(int j=0; j<i; j++) //分別選取前 i 個醜數乘2、3、5，所得結果最小作爲對應值
{
if(v[i-1] < v[j]*2)
v[i] = min(v[i], v[j]*2);
if(v[i-1] < v[j]*3)
v[i] = min(v[i], v[j]*3);
if(v[i-1] < v[j]*5)
v[i] = min(v[i], v[j]*5);
}
return v[index-1];
}
private:
vector<int> v;
};
優化：
class Solution {
public:
int GetUglyNumber_Solution(int index) {
if(index == 0) //不加這個報錯！
return 0;
if(index == 1)
return 1;
v = vector<int>(index, 0x3f3f3f3f);
v[0] = 1;
int pos2 = 0;
int pos3 = 0;
int pos5 = 0;
for(int i=1; i<index; i++) //計算第 i+1 個醜數
{
v[i] = min(v[pos2]*2, min(v[pos3]*3, v[pos5]*5));
if(v[i] == v[pos2]*2)
pos2++;
if(v[i] == v[pos3]*3)
pos3++;
if(v[i] == v[pos5]*5)
pos5++;
}
return v[index-1];
}
private:
vector<int> v;
};
34. 第一個只出現一次的字符位置
class Solution {
public:
int FirstNotRepeatingChar(string str) {
for(int i=0; i<str.length(); i++)
{
m[str[i]]++;
}
for(int i=0; i<str.length(); i++)
if(m[str[i]] == 1)
return i;
return -1;
}
private:
map<char, int> m;
};
35. 兩個鏈表的第一個公共結點
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {\
set<ListNode *> s;
for (auto it = pHead1; it != nullptr; it = it->next)
s.insert(it);
for (auto it = pHead2; it != nullptr; it = it->next)
if (s.find(it) != s.end())
return it;
return nullptr;
}
};
數組長度
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {\
ListNode *p1 = pHead1;
int len1 = 0;
ListNode *p2 = pHead2;
int len2 = 0;
while(p1)
{
p1 = p1->next;
len1++;
}
while(p2)
{
p2 = p2->next;
len2++;
}
p1 = pHead1;
p2 = pHead2;
if(len1 > len2)
{
int dis = len1-len2;
while(dis-->0)
p1 = p1->next;
}else{
int dis = len2-len1;
while(dis-->0)
p2 = p2->next;
}
while(p1 && p2)
{
if(p1 == p2)
return p1;
p1 = p1->next;
p2 = p2->next;
}
return nullptr;
}
};
36. 數字在排序數組中出現的次數
class Solution {
public:
int GetNumberOfK(vector<int> data ,int k) {
int ret = 0;
int l = 0;
int r = data.size()-1;
while(l <= r)
{
int mid = l+(r-l)/2;
if(data[mid] == k)
{
int pre = mid-1;
while(pre>=0 && data[pre]==k)
{
ret++;
pre--;
}
int post = mid+1;
while(post<data.size() && data[post]==k)
{
ret++;
post++;
}
return ret+1;
}else if(data[mid] > k){
r = mid-1;
}else{
l = mid+1;
}
}
return 0;
}
};
37. 二叉樹的深度
/*
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)
{
int ret = 0;
if(!pRoot)
return ret;
inOrder(pRoot, 1, ret);
return ret;
}
private:
void inOrder(TreeNode *node, int level, int &ret)
{
if(!node)
return ;
if(node->left)
inOrder(node->left, level+1, ret);
ret = max(ret, level);
if(node->right)
inOrder(node->right, level+1, ret);
}
};
精簡版
/*
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 == nullptr)
return 0;
return max(TreeDepth(pRoot->left), TreeDepth(pRoot->right)) + 1;
}
};
38. 平衡二叉樹
class Solution {
public:
bool IsBalanced_Solution(TreeNode* pRoot) {
if(!pRoot)
return true;
return preorder(pRoot);
}
private:
int treeDepth(TreeNode *node)
{
if(!node)
return 0;
return max(treeDepth(node->left), treeDepth(node->right))+1;
}
bool preorder(TreeNode *node)
{
if(!node)
return true;
int lDepth = treeDepth(node->left);
int rDepth = treeDepth(node->right);
return abs(lDepth-rDepth)<=1&&preorder(node->left)&&preorder(node->right);
}
};
39. 數組中只出現一次的數字
- 數組異或
- 找出結果中的最低一位爲1
- 按照這一位是否爲1把數組分爲兩組
- 每組分別各自異或
class Solution {
public:
void FindNumsAppearOnce(vector<int> data,int* num1,int *num2) {
if(data.size() < 2)
return ;
int tmp = data[0];
for(int i=1; i<data.size(); i++)
tmp ^= data[i];
//找到最低位爲1
int index = 0;
while((tmp&1) == 0)
{
tmp = tmp >> 1;
index++;
}
int n1 = 0;
int n2 = 0;
for(int i=0; i<data.size(); i++)
{
//判斷第 index 位是否爲 1
if(isBit(data[i], index))
n1 ^= data[i];
else
n2 ^= data[i];
}
*num1 = n1;
*num2 = n2;
}
private:
bool isBit(int num, int index)
{
if(num>>index & 1 != 0)
return true;
return false;
}
};
40. 和爲S的連續正數序列
雙指針
class Solution {
public:
vector<vector<int> > FindContinuousSequence(int sum) {
vector<vector<int>> ret;
if(sum < 3)
return ret;
int l = 1;
int r = 2;
while(l < r)
{
int n = (l+r)*(r-l+1)/2; //求出 [l...r] 的和
if(n == sum)
{
vector<int> v;
for(int i=l; i<=r; i++)
v.push_back(i);
ret.push_back(v);
l++;
}else if(n < sum){
r++;
}else
l++;
}
return ret;
}
};