算法题整理
作者：互联网
1.最长回文子串
class Solution {
public:
string isPalindrome(string& s, int left, int right) {
while (left >= 0 && right <= s.length() - 1 && s[left] == s[right]) {
left --;
right ++;
}
return s.substr(left+1, right-left-1);
}
string longestPalindrome(string s) {
if (s.empty()) return s;
int left = 0, right = 0;
string longestpld = "";
for (int start = 0; start < s.length(); start ++) {
string str = isPalindrome (s, start,start);
if (longestpld.length() < str.length()) longestpld = str;
str = isPalindrome (s, start,start + 1);
if (longestpld.length() < str.length()) longestpld = str;
}
return longestpld;
}
};
2. 快速排序
void quicksort(int *a, int start, int end) {
if (start >= end) {
return;
}
int right = end;
int left = start;
int middle = a[(right+left)/2];
while (left <= right) {
while (left <= right && a[left] < middle)
left++;
while (left <= right && a[right] > middle)
right--;
if (left <= right) {
int temp = a[left];
a[left] = a[right];
a[right] = temp;
left++;
right--;
}
}
quicksort (a, start, right);
quicksort (a, left, end);
}
3. 归并排序
#include <iostream>
using namespace std;
void merge(int *a, int start, int end, int *temp) {
int middle = (start + end)/2;
int left = start;
int right = middle +1;
int index = left;
while(left <= middle && right <= end) {
if (a[left] <= a[right]){
temp[index++] = a[left++];
}
else {
temp[index++] = a[right++];
}
}
while (left <= middle) {
temp[index++] = a[left++];
}
while (right <= end) {
temp[index++] = a[right++];
}
for ( int i = start; i<= end; i++){
a[i] = temp[i];
}
}
void mergesort(int *a, int start, int end, int *temp) {
if (start >= end) {
return;
}
mergesort(a, start, (start+end)/2, temp);
mergesort(a, (start+end)/2 + 1, end, temp);
merge(a, start, end, temp);
}
int main(int argc, char* argv[]){
int a[] = {2,8,4,10,3,5,7,9};
const int num = sizeof(a)/sizeof(a[0]);
int b[num];
mergesort(a, 0, num - 1, b);
for (int i = 0; i< sizeof(a)/sizeof(a[0]); i++){
cout << a[i] <<endl;
}
}
4. 快速选择排序
#include <iostream>
using namespace std;
int quickselect(int *a, int start, int end, int k) {
if (start >= end) {
return a[start];
}
int left = start;
int right = end;
int middle = a[(start+end)/2];
while (left <= right) {
while (left <= right && a[left] > middle)
left++;
while (left <= right && middle > a[right])
right--;
if (left <= right) {
int temp = a[left];
a[left] = a[right];
a[right] = temp;
left++;
right--;
}
}
if ( start +k -1 <= right) {
return quickselect (a, start, right, k);
}
else if(start +k -1 >= left) {
return quickselect (a, left, end, k - (left - start)) ;
}
return a[right+1];
}
int main(int argc, char* argv[]){
int a[] = {2,8,4,10,3,5,7,9};
const int num = sizeof(a)/sizeof(a[0]);
int k = 3;
int n = quickselect(a, 0, num - 1, k);
cout << n << endl;
}
5.二分法
三数之和
class Solution {
public:
void towsum (vector<int>& nums, int index, vector<vector<int>>& results) {
int start = index + 1;
int end = nums.size()-1;
int target = -nums[index];
while(start < end) {
if (nums[start] + nums[end] > target) {
end--;
}
else if (nums[start] + nums[end] < target) {
start++;
}
else {
results.push_back(vector<int>{-target, nums[start],nums[end]});
start++;
end--;
while (start < end && nums[start] == nums[start-1]){
start++;
}
}
}
}
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> results;
if (nums.size()<3){
return results;
}
sort(nums.begin(), nums.end());
for (int i = 0; i<nums.size()-2; i++){
if (i!=0 && nums[i] == nums[i-1])
continue;
towsum(nums, i, results);
}
return results;
}
};
三角形个数
class Solution {
public:
/**
* @param s: A list of integers
* @return: An integer
*/
int getnum(vector<int> &s, int index){
int left = 0;
int right = index-1;
int ans = 0;
while (left < right){
if (s[left] + s[right] > s[index]) {
ans += right - left;
right--;
}
else {
left++;
}
}
return ans;
}
int triangleCount(vector<int> &s) {
if (s.size()<3) {
return 0;
}
sort(s.begin(), s.end());
int ans = 0;
for (int i = 2; i<s.size(); i++) {
ans += getnum(s, i);
}
return ans;
}
};
四数之和
class Solution {
public:
/**
* @param numbers: Give an array
* @param target: An integer
* @return: Find all unique quadruplets in the array which gives the sum of zero
* we will sort your return value in output
*/
void twosum (vector<int> &numbers, int index1, int index2, long target, vector<vector<int>>& res) {
int left = index2 + 1;
int right = numbers.size() - 1;
while(left < right) {
if (numbers[left] + numbers[right] > target) {
right--;
}
else if (numbers[left] + numbers[right] < target) {
left++;
}
else {
res.push_back(vector<int> {numbers[index1], numbers[index2], numbers[left], numbers[right]});
left++;
right--;
while (left < right && numbers[left] == numbers[left-1]){
left++;
}
}
}
}
vector<vector<int>> fourSum(vector<int> &numbers, int target) {
vector<vector<int>> res;
if (numbers.size()<4){
return res;
}
sort (numbers.begin(), numbers.end());
for (int i = 0; i<numbers.size()-3; i++){
if (i != 0 && numbers[i] == numbers[i-1])
continue;
for (int j = i+1; j<numbers.size()-2; j++){
if (j != i+1 && numbers[j] == numbers[j-1])
continue;
twosum(numbers, i, j, (long)target-(numbers[i] + numbers[j]), res);
}
}
return res;
}
};
四数组的四数之和
class Solution {
public:
/**
* @param a: a list
* @param b: a list
* @param c: a list
* @param d: a list
* @return: how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero
*/
int fourSumCount(vector<int> &a, vector<int> &b, vector<int> &c, vector<int> &d) {
if (!a.size() || !b.size() || !c.size() || !d.size()){
return 0;
}
unordered_map <int, int> hashTable;
int res = 0;
for (int i = 0; i < a.size(); i++){
for (int j = 0; j < b.size(); j++){
hashTable[a[i]+b[j]]++;
}
}
for (int p = 0; p < c.size(); p++){
for (int q = 0; q < d.size(); q++){
int num = c[p] + d[q];
if (hashTable.find(-num) != hashTable.end()){
res += hashTable[-num];
}
}
}
return res;
}
};
标签：right,end,int,start,算法,vector,整理,left 来源： https://www.cnblogs.com/cuijy1/p/16637844.html