leetcoders / leetcode

Solve problems from LeetCode.

LeetCode

Solve problems from Leetcode. All the codes are tested using online-judge.

Here is a difficulty and frequency distribution chart for each problem, which I got from the Internet and is very useful. Feel free to make pull request for adding the difficulty and frequency for new problems here.

Please feel free to let me know if you have any problem or better solutions:)

Note: The problem link in the file will take you to the Old OJ, which will expire on October 12, 12pm PST.
You may use this link to the new OJ instead!

leetcode's Issues

scaramble string : my dp solution

首先想了个不到n^3的dp，后来发现不适合有字符重复的情况，下面这个dp解法能过，不过我觉得时间复杂度有点高了
sub[len][i][j] 表示s1从i开始与s2从j开始的长度为len是否为Scramble String

class Solution {
public:
bool sub[100][100][100];
bool isScramble(string s1, string s2) {
if(s1.length()!=s2.length())
return false;
int len = s1.length();
for(int i = 0;i<len;i++)
for(int j = 0;j<len;j++)
sub[1][i][j] = (s1[i]==s2[j])?1:0;
for(int k = 2;k<=len;k++)
for(int i = 0;i<=len- k ;i++)
for(int j = 0;j<=len-k;j++)
{
sub[k][i][j] = 0;
for(int p =i+1;p<i+k&&!sub[k][i][j];p++)
if( sub[p-i][i][j] && sub[k-p+i][p][p-i+j] )
sub[k][i][j] =1;
else if( sub[p-i][i][k-p+i+j] &&sub[k-p+i][p][j])
sub[k][i][j]=1;
else
sub[k][i][j] = 0;
}
return sub[len][0][0];

}

};

subsets:非递归版本

class Solution {
public:
vector<vector<int> > subsets(vector<int> &S) {
    int size = S.size();
    vector<vector<int>> result;
    if(size==0) return result;
    sort(S.begin(),S.end());
    vector<int>v;
    result.push_back(v);
    for(int i=1;i<=size;i++)
    {
        int*p = new int[i];
        p[0]=0;
        int index = 0;
        while(1)
        {
            if(p[index] == size)
            {
                if(index==0)break;
                index--;
                p[index]++;
            }
            else if(index == i-1)
            {
                v.clear();
                for(int j = 0;j<i;j++)
                    v.push_back(S[p[j]]);
                result.push_back(v);
                p[index]++;
            }
            else
            {
                index++;
                p[index]=p[index-1]+1;
            }

        }
        delete[]p;
        p =NULL;
    }

    return result;

}
};

link for distribution chart is invalid

The 'difficulty and frequency distribution chart' link (https://docs.google.com/spreadsheet/pub?key=0Aqt--%20wSNYfuxdGxQWVFsOGdVVWxQRlNUVXZTdEpOeEE&output=html) described in README.md is invalid. Does anybody have a backup of it?

bug in "Interleaving String"

In such case: _s1 = "b", s2 = "acca", s3 = "abcba"_, it would return true.

This is related to:

    for (int i = 1; i <= N; ++i)
        dp[i][0] = s2[i-1] == s3[i-1];
    for (int j = 1; j <= M; ++j)
        dp[0][j] = s1[j-1] == s3[j-1];

Binary Tree Zigzag Level Order Traversal

class Solution {
public:
vector<vector > zigzagLevelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector> result;
if(root==NULL) return result;
vector<TreeNode * >v;
v.push_back(root);
bool leftsonfirst = true;
int index,begin = 0;
while(1)
{
vector newv;
newv.clear();
index = v.size();
if(index == begin) break;
int p = index-1;
while(p>=begin)
{
if(v[p]!=NULL)
{
newv.push_back(v[p]->val);
if(leftsonfirst)
{
v.push_back(v[p]->left);
v.push_back(v[p]->right);
}
else
{
v.push_back(v[p]->right);
v.push_back(v[p]->left);
}
}
p--;
}
if(newv.size())
result.push_back(newv);
begin = index;
leftsonfirst = !leftsonfirst;
}
return result;
}

};