目录
1.回文数
https://leetcode.cn/problems/palindrome-number/description/
cpp
class Solution {
public:
bool isPalindrome(int x) {
//当x<0时肯定不为回文数,最后一位为0,若要回文只有0满足
if(x < 0 || x % 10 == 0 && x != 0)
return false;
int ReverteNumber = 0;
while(x > ReverteNumber)
{
ReverteNumber = ReverteNumber * 10 + x % 10;
x /= 10;
}
return x == ReverteNumber || x == ReverteNumber / 10;
}
};2.盛最多水的容器
https://leetcode.cn/problems/container-with-most-water/description/
cpp
class Solution {
public:
int maxArea(vector<int>& height) {
//双指针,left左指针,right右指针
int left = 0,right = height.size()-1;
int ans=0;
while(left<right)
{
//面积为左右指针值小的*指针距离
int area = min(height[left],height[right])*(right-left);
//更新ans的值,存储面积最大的
ans=max(ans,area);
//每次移动值较小的指针
if(height[left]<=height[right])
++left;
else
--right;
}
return ans;
}
};3.罗马数字转整数
https://leetcode.cn/problems/roman-to-integer/
cpp
int romanToInt(char* s) {
int symbolValues[26];
//用ASII值相减,symbolValues[8]用来表示I,即数字1
symbolValues['I' - 'A'] = 1;
symbolValues['V' - 'A'] = 5;
symbolValues['X' - 'A'] = 10;
symbolValues['L' - 'A'] = 50;
symbolValues['C' - 'A'] = 100;
symbolValues['D' - 'A'] = 500;
symbolValues['M' - 'A'] = 1000;
int ans = 0;
int n = strlen(s);
for(int i = 0;i<n;++i)
{
int value = symbolValues[s[i]-'A'];
if(i < n-1 && value < symbolValues[s[i+1] - 'A'])
{
ans -= value;
}
else
{
ans += value;
}
}
return ans;
}4.最长公共前缀
https://leetcode.cn/problems/longest-common-prefix/
cpp
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
if(strs.size() == 0)
return "";
int length = strs[0].size();
int count = strs.size();
for(int i = 0;i<length;++i)//遍历第一个字符串
{
char c = strs[0][i];
for(int j = 1;j<count;++j)
{
if(i == strs[j].size() || strs[j][i] != c)
{
//substr从某一位置提取一定字符
return strs[0].substr(0,i);
}
}
}
return strs[0];
}
};5.三数之和
https://leetcode.cn/problems/3sum/description/
cpp
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
// ranges::sort(nums);
//数组排序
sort(nums.begin(),nums.end());
vector<vector<int>> ans;
int n = nums.size();
// 遍历数组,将每个元素作为可能的三元组中的第一个元素
for (int i = 0; i < n - 2; i++) {
int x = nums[i];// 当前元素
if (i && x == nums[i - 1]) continue; // 跳过重复数字
if (x + nums[i + 1] + nums[i + 2] > 0) break; // 当前元素与后两个之和大于0,结束循环
if (x + nums[n - 2] + nums[n - 1] < 0) continue; // 当前元素与后两个之和小于0,跳过当前循环
int j = i + 1, k = n - 1;// 双指针找剩余两个数,i的后一位和数组的最后一位
while (j < k) {
int s = x + nums[j] + nums[k];
if (s > 0) {
k--;
} else if (s < 0) {
j++;
} else {
ans.push_back({x, nums[j], nums[k]});// 找到符合条件的加入列表
for (j++; j < k && nums[j] == nums[j - 1]; j++); // 跳过重复数字
for (k--; k > j && nums[k] == nums[k + 1]; k--); // 跳过重复数字
}
}
}
return ans;
}
};6.最接近的三数之和
https://leetcode.cn/problems/3sum-closest/description/
cpp
class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
sort(nums.begin(), nums.end());
int ans,n = nums.size();
int min_diff = INT_MAX;// 最大正整数
for(int i = 0; i < n - 2; i++){
int x = nums[i];
if(i > 0 && x == nums[i - 1]) continue;
int s = x + nums[i + 1] + nums[i + 2];
if(s > target){
if(s - target < min_diff){
ans = s;
}
break;
}
s = x + nums[n - 2] + nums[n - 1];
if(s < target){
if(target - s < min_diff){
min_diff = target - s;
ans = s;
}
continue;
}
int j = i + 1, k = n - 1;
while(j < k){
s = x + nums[j] + nums[k];
if(s == target){
return target;
}
if(s > target){
if(s - target < min_diff){
min_diff = s - target;
ans = s;
}
k--;
}
else{
if(target - s < min_diff){
min_diff = target - s;
ans = s;
}
j++;
}
}
}
return ans;
}
};