文章目录
61.扑克牌中的顺子
//java版
class Solution {
public boolean isStraight(int[] nums) {
if(nums.length != 5) return false;
Arrays.sort(nums);// 0 0 xxx
//去除0
int k = 0;
while(nums[k] == 0){
//2
k ++; //3
}
//nums[k] != 0
//看有没有重复数字 如果有 false
for(int i = k + 1; i < nums.length; i ++)
if(nums[i] == nums[i - 1])
return false;
//看除了0以外的最大 - 最小 <= 4
return nums[4] - nums[k] <= 4;
}
}
//C++版
class Solution {
public:
bool IsContinuous( vector<int> nums ) {
if(nums.empty()) return false;
//大小王 0
//去除0
sort(nums.begin(), nums.end());
int k = 0;
while(nums[k] == 0) k ++;
//找从k + 1位开始的 有没有重复数字
for(int i = k + 1; i < nums.size(); i ++)
{
if(nums[i] == nums[i - 1])
return false;
}
//确保最大值-最小值 <= 4
return nums.back() - nums[k] <= 4;
}
};
62.圆圈中最后剩下的数字
//java版
class Solution {
public int lastRemaining(int n, int m) {
//旧编号 0 1 2 ... m-1 m m+1 ... n-1
//新编号 -m -m+1 -m+2 ... x 0 1 ... n-1-m
//递推关系 f(n,m) = (f(n - 1, m) + m) % n
if(n == 1) return 0;
return (lastRemaining(n - 1, m) + m) % n;
}
}
//C++版
class Solution {
public:
int LastRemaining_Solution(int n, int m)
{
if(n == 0) return -1;
if(n == 1) return 0;
return (LastRemaining_Solution(n - 1, m ) + m) % n;
}
};
63.股票的最大利润
//java版
class Solution {
public int maxProfit(int[] prices) {
if(prices.length == 0)
return 0;
int maxv = 0;//最大利润
int minv = prices[0];//最小价格，默认第一个价格最小
for(int i = 1; i < prices.length; i ++){
maxv = Math.max(prices[i] - minv, maxv);
minv = Math.min(minv, prices[i]);
}
return maxv;
}
}
//C++版
class Solution {
public:
int maxProfit(vector<int>& nums) {
if(nums.empty()) return 0;
int maxv = 0;
for(int i = 1, minv = nums[0]; i < nums.size(); i ++)
{
maxv = max(maxv, nums[i] - minv);
minv = min(minv, nums[i]);
}
return maxv;
}
};
64.求1+2+…+n
//java版
class Solution {
int[] test=new int[]{
0};
public int sumNums(int n) {
try{
return test[n];
}catch(Exception e){
return n+sumNums(n-1);
}
}
}
//C++版
class Solution {
public:
int Sum_Solution(int n) {
int res = n;
(n > 0) && (res += Sum_Solution(n - 1));
return res;
}
};
65.不用加减乘除做加法
//java版
class Solution {
public int add(int a, int b) {
//与 异或
while(b != 0){
//进位找到0就停止 a就是和
int c = (a & b) << 1;//c进位
a ^= b;//a非进位部分的和
b = c;
}
return a;
}
}
//C++版
class Solution {
public:
int Add(int num1, int num2)
{
// & ^
//^ 相同 0 不同1
//0^0 1 1^1 0 0^ 1 1
//利用这个性质，我们能够得到32位2进制中 不进1的位置
//1 0^0 0^1
//找到进1的位置
// 1 & 1 = 1
while(num2)
{
int sum = num1 ^ num2;
int carry = (unsigned int)(num1 & num2) << 1;
num1 = sum;
num2 = carry;
}
//out = sum + carry
return num1;
}
};
66.构建乘积数组
//java版
class Solution {
public int[] constructArr(int[] a) {
//想办法在 i的时候 避开× ai
int n = a.length;
int[] res = new int[n];
for(int i = 0,p = 1; i < n; i ++){
res[i] = p;// a1 **** ai -1
p *= a[i];
}
for(int i = n - 1, p = 1; i >= 0; i --){
res[i] *= p;//ai+1 **** an-1
p *= a[i];
}
return res;
}
}
//C++版
class Solution {
public:
vector<int> multiply(const vector<int>& A) {
if(A.empty()) return vector<int>();
int n = A.size();
vector<int> B(n);
// i 左边
for(int i = 0, p = 1; i < n; i ++)
{
//a1 *...* ai-1
B[i] = p;
p *= A[i];
}
//i右边
//ai+1 *....* an-1
for(int i = n - 1,p = 1; i >= 0; i --)
{
B[i] *= p;
p *= A[i];
}
return B;
}
};
67.把字符串转换成整数
//java版
class Solution {
public int strToInt(String str) {
//2147483647 214748364 + 还有一个数字 7比较大小
//-2147483648 214748364 + 还有一个数字
char[] c = str.trim().toCharArray();
if(c.length == 0)
return 0;
int res = 0, sign = 1;//res返回结果 sign 1正数 -1负数
int temp = Integer.MAX_VALUE / 10;//214748364
int index = 1;
if(c[0] == '-') sign = -1;
else if(c[0] != '+') index = 0;
for(int i = index; i < c.length; i ++){
//"4193 with words" 4193
//"words and 987" 0
if(c[i] < '0' || c[i] > '9') break;
if(res > temp || (res == temp && c[i] > '7'))
return sign == 1 Integer.MAX_VALUE : Integer.MIN_VALUE;
res = res * 10 + c[i] - '0';
}
return sign * (int)res;
}
}
//C++版
class Solution {
public:
int StrToInt(string str) {
int k = 0;
long long num = 0;
int sign = 1;
if(str[k] == '-') k ++, sign = -1;
else if(str[k] == '+') k ++, sign = 1;
while(k < str.size())
{
if(str[k] >= '0' && str[k] <= '9')
{
num = num * 10 + str[k] - '0';
k ++;
}
else{
num = 0;
break;
}
}
num *= sign;
return num;
}
};
68.二叉搜索树的最近公共祖先
//java版
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(p.val > q.val){
//确保p < q
TreeNode temp = p;
p = q;
q = temp;
}
while(root != null){
if(root.val < p.val)
root = root.right;
else if(root.val > q.val)
root = root.left;
else break;
}
return root;
}
}
//C++版
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(!root) return NULL;
if(root == p || root == q) return root;
auto left = lowestCommonAncestor(root -> left, p, q);
auto right = lowestCommonAncestor(root -> right, p, q);
if(left && right) return root;
if(left) return left;
return right;
}
};
二叉树的最近公共祖先
//java版
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root == null || root == p || root == q) return root;
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if(left == null) return right;
if(right == null) return left;
return root;
}
}