leetcode习题代码(经常更新)

为了准备面试编程题，我最近准备刷一刷leetcode，我是从191开始往前刷，把accept过的代码贴上来

leetcode 191

class Solution {
public:
int hammingWeight(uint32_t n) {
int ans = 0;
long long tmp = 1;
while(tmp <= n)
{
if ((tmp & n) != 0)
{
ans ++;
}
//            printf("tmp = %d tmp & n = %d\n",tmp,tmp & n);
tmp <<= 1;
}
return ans;
}
};

leetcode 190:

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
long long ans = 0;
long long tmp = 1;
for (unsigned int i=1; i;i<<=1)
{
ans <<= 1;
if (tmp & n)
{
ans ++;
}
tmp <<= 1;
}
return ans;
}
};

189没看明白题目。不过看起来很简单的样子

leetcode 188，说实话，这题是我见过最奇怪的动态规划，像背包，但是又想不出好的解法，一时半会儿真弄不明白。有篇博客讲得不错：http://blog.csdn.net/linhuanmars/article/details/23236995

class Solution
{
public:
int maxProfit(int k, vector<int> &prices)
{
int size = prices.size();
if (k >= size)
{
int sum = 0;
for (int i=1; i<size; i++)
{
sum += max(prices[i] - prices[i - 1],0);
}
return sum;
}
int local[k + 1];
int global[k + 1];
for (int i=0; i<=k; i++)
{
local[i] = global[i] = 0;
}
for (int i=0; i<size - 1; i++)
{
int length = prices[i + 1] - prices[i];
for (int j=k; j>=1; j--)
{
local[j] = max(global[j - 1] + max(length,0),local[j] + length);
global[j] = max(global[j],local[j]);
}
}
return global[k];
}
};

leetcode 187

这题坑爹的地方，我用hash居然报MLE，我把数组开到最小居然还报，后来改成map才过，这烂judge真坑爹！

下面是hash+vector版本:

class Solution {
public:

vector<string> f[10000];

int get_num(char c)
{
int ans = 0;
switch (c)
{
case 'A':
{
ans = 1;
break;
}
case 'G':
{
ans = 2;
break;
}
case 'C':
{
ans = 3;
break;
}
case 'T':
{
ans = 4;
break;
}
}
return ans;
}

long long get_sum(int num)
{
long long ans = num;
for (int i=0; i<9; i++)
{
ans *= 10;
}
return ans;
}

vector<string> findRepeatedDnaSequences(string s)
{
vector<string> ans;
int size = s.length();

int mod = 9997;
long long sum = 0;
long long sum1 = 0;
int start = 0,tail = 10;
if(s == "" || size == 0)
{
printf("size = %d\n",size);
return ans;
}
for (int i=0; i<10; i++)
{
int tmp = get_num(s[i]);
sum1 *= 10;
sum1 += tmp;
sum = sum1 % mod;
}
if (size > tail)
{
f[sum].push_back(s.substr(0,10));;
start = 1;
while(tail < size)
{
sum1 -= (get_sum(get_num(s[start - 1])));;
sum1 *= 10;
sum1 += get_num(s[tail]);
sum = sum1 % mod;
if (f[sum].size() > 0)
{
for (int i=0; i<f[sum].size(); i++)
{
if (f[sum][i] == s.substr(start,10))
{
ans.push_back(s.substr(start,10));
}
else
{
f[sum].push_back(s.substr(start,10));
}
}
}
else
{
f[sum].push_back(s.substr(start,10));
}
start ++;
tail ++;
}
}
return ans;
}
};//mle

AC版本map:

int get_num(char c)
{
int ans = 0;
switch (c)
{
case 'A':
{
ans = 1;
break;
}
case 'G':
{
ans = 2;
break;
}
case 'C':
{
ans = 3;
break;
}
case 'T':
{
ans = 4;
break;
}
}
return ans;
}

long long get_sum(int num)
{
long long ans = num;
for (int i=0; i<9; i++)
{
ans *= 10;
}
return ans;
}

vector<string> findRepeatedDnaSequences(string s)
{
vector<string> ans;
int size = s.length();

int mod = 990023;
long long sum = 0;
long long sum1 = 0;
map<long long,int> m;
int start = 0,tail = 10;
if(s == "" || size == 0)
{
return ans;
}
for (int i=0; i<10; i++)
{
int tmp = get_num(s[i]);
sum1 *= 10;
sum1 += tmp;
}
m[sum1] ++;
if (size > tail)
{
start = 1;
while(tail < size)
{
sum1 -= (get_sum(get_num(s[start - 1])));;
sum1 *= 10;
sum1 += get_num(s[tail]);
if (m[sum1] > 0)
{
if (m[sum1] == 1)
ans.push_back(s.substr(start,10));
}
m[sum1]++;
start ++;
tail ++;
}
}
//    for (int i=0; i<ans.size(); i++)
//    {
//        cout<<ans[i]<<endl;
//    }
return ans;
}

两种解法时间复杂度都是O(n);

leetcode 179

这题有两种方案，一是搜索，但无疑会Tle，另一种是排序，两个方案的思想都很好，不针对题目，将两个方案的代码都列出来:

1.搜索代码TLE:

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;

string int2str(int num)
{
int tmp = 0;
string ans;
char s[100];
int length = 0;
while(num > 0)
{
tmp = num % 10;
s[length++] = (tmp + '0');
num /= 10;
}
s[length] = '\0';
reverse(s,s+length);
ans = s;
//    cout<<"ans = "<<ans<<endl;
return ans;
}
string ans;

bool vis[10010];

void get_ans(vector<int> &num,string tmp,int n)
{
int size = num.size();
if (n == num.size() - 1)
{
if (ans < tmp)
{
ans = tmp;
}
return ;
}
for (int i=0; i<size; i++)
{
if (vis[i] == 0)
{
vis[i] = 1;
get_ans(num,tmp + int2str(num[i]),n + 1);
vis[i] = 0;
}
}
}

string largestNumber(vector<int> &num)
{
get_ans(num,"",0);
return ans;
}

int main()
{
memset(vis,0,sizeof(vis));
vector<int> num;
for (int i=0; i<6; i++)
{
num.push_back(i);
}
largestNumber(num);
cout<<"ans = " <<ans<<endl;
return 0;
}

2。java排序方法

public class Solution {
public static String largestNumber(int[] num) {
int length = num.length;
String[] ans = new String[length];
for (int i=0; i<length; i++)
{
ans[i] = num[i] + "";
}
Arrays.sort(ans,new MyCompare());
String res = "";
for (int i=0; i<length; i++)
{
res += ans[i];
}	int index = -1;
for (int i=0; i<res.length(); i++)
{
if (res.charAt(i) == '0')
{
index = i;
}
else
{
break;
}
}
if (index != -1)
return res.substring(index);
else {
return res;
}
}
}

class MyCompare implements Comparator<String>
{

public int compare(String o1, String o2) {
String s1 = (String) o1;
String s2 = (String) o2;
int res = (s1 + s2).compareTo(s2 + s1);
if (res < 0)
return 1;
else {
return -1;
}
}

}

leetcode 174比较简单的dp

public int calculateMinimumHP(int[][] dungeon) {
int m = dungeon.length;
int n = dungeon[0].length;

int[][] ans = new int[m]
;

ans[m - 1][n - 1] = dungeon[m - 1][n - 1] > 0 ? 0 : -dungeon[m - 1][n - 1];
for (int i = m - 2; i >= 0; i--) {
ans[i][n - 1] = dungeon[i][n - 1] >= ans[i + 1][n - 1] ? 0 : ans[i + 1][n - 1] - dungeon[i][n - 1];
}
for (int j = n - 2; j >= 0; j--) {
ans[m - 1][j] = dungeon[m - 1][j] >= ans[m - 1][j + 1] ? 0 : ans[m - 1][j + 1] - dungeon[m - 1][j];
}

for (int i = m - 2; i >= 0; i--) {
for (int j = n - 2; j >= 0; j--) {
int down = dungeon[i][j] >= ans[i + 1][j] ? 0 : ans[i + 1][j] - dungeon[i][j];
int right = dungeon[i][j] >= ans[i][j + 1] ? 0 : ans[i][j + 1] - dungeon[i][j];
ans[i][j] = Math.min(down, right);
}
}

return ans[0][0] + 1;
}