柳婼の blog – 第244页 – 我不管，反正我最萌～

LeetCode 155. Min Stack

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

push(x) — Push element x onto stack.
pop() — Removes the element on top of the stack.
top() — Get the top element.
getMin() — Retrieve the minimum element in the stack.

class MinStack {
public:
    stack<int> s;
    stack<int> min;
    
    void push(int x) {
        s.push(x);
        if(min.empty() || min.top() >= x) {
            min.push(x);
        }
    }//这里 if 语句里面是 >= 的原因是，假设有两个相同的最小值8，
    //如果只存一个（只填 > ），假设两个 stack 同时pop了一个8，
    //那么其实最小值还有8但是已经被 pop 掉了

    void pop() {
        if(s.top() == min.top()) {
            s.pop();
            min.pop();
        } else {
            s.pop();
        }
    }

    int top() {
        return s.top();
    }

    int getMin() {
        return min.top();
    }
};

class MinStack {

public:

stack<int> s;

stack<int> min;

void push(int x) {

s.push(x);

if(min.empty() || min.top() >= x) {

min.push(x);

}

}//这里 if 语句里面是 >= 的原因是，假设有两个相同的最小值8，

//如果只存一个（只填 > ），假设两个 stack 同时pop了一个8，

//那么其实最小值还有8但是已经被 pop 掉了

void pop() {

if(s.top() == min.top()) {

s.pop();

min.pop();

} else {

s.pop();

}

int top() {

return s.top();

}

int getMin() {

return min.top();

}

};

LeetCode 34. Search for a Range

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int low = 0, high = nums.size() - 1;
        while(low <= high && nums[low] != target && nums[high] != target) {
            int mid = (high - low) / 2 + low;
            if(nums[mid] < target) {
                low = mid + 1;
            } else if(nums[mid] > target){
                high = mid - 1;
            } else {
                break;
            }
        }
        vector<int> v(2);
        if(low > high) {
            v[0] = -1;
            v[1] = -1;
            return v;
        }
        while(nums[low] != target) {
            low++;
        }
        while(nums[high] != target) {
            high--;
        }
        v[0] = low;
        v[1] = high;
        return v;
    }
};

class Solution {

public:

vector<int> searchRange(vector<int>& nums, int target) {

int low = 0, high = nums.size() - 1;

while(low <= high && nums[low] != target && nums[high] != target) {

int mid = (high - low) / 2 + low;

if(nums[mid] < target) {

low = mid + 1;

} else if(nums[mid] > target){

high = mid - 1;

} else {

break;

}

vector<int> v(2);

if(low > high) {

v[0] = -1;

v[1] = -1;

return v;

}

while(nums[low] != target) {

low++;

}

while(nums[high] != target) {

high--;

}

v[0] = low;

v[1] = high;

return v;

}

};

LeetCode 278. First Bad Version

You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.

Suppose you have n versions [1, 2, …, n] and you want to find out the first bad one, which causes all the following ones to be bad.

You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int low = 1, high = n;
        while(low <= high) {
            int mid = (high - low) / 2 + low;
            if(isBadVersion(mid)) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }
};

// Forward declaration of isBadVersion API.

bool isBadVersion(int version);

class Solution {

public:

int firstBadVersion(int n) {

int low = 1, high = n;

while(low <= high) {

int mid = (high - low) / 2 + low;

if(isBadVersion(mid)) {

high = mid - 1;

} else {

low = mid + 1;

}

return low;

}

};

LeetCode 7. Reverse Integer

Example1: x = 123, return 321
Example2: x = -123, return -321

Have you thought about this?
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer’s last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

class Solution {
public:
    int reverse(int x) {
        string s = to_string(x);
        if(s[0] == '-')
            std::reverse(s.begin() + 1, s.end());
        else
            std::reverse(s.begin(), s.end());
        long long int temp = stoll(s);
        if(temp > 2147483647 || temp < -2147483648)
            return 0;
        return (int)temp;
    }
};

class Solution {

public:

int reverse(int x) {

string s = to_string(x);

if(s[0] == '-')

std::reverse(s.begin() + 1, s.end());

else

std::reverse(s.begin(), s.end());

long long int temp = stoll(s);

if(temp > 2147483647 || temp < -2147483648)

return 0;

return (int)temp;

}

};

LeetCode 125. Valid Palindrome

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,
“A man, a plan, a canal: Panama” is a palindrome.
“race a car” is not a palindrome.

Note:
Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.

class Solution {
public:
    bool isPalindrome(string s) {
        if(s.length() == 0)
            return true;
        int i = 0, j = s.length() - 1;
        while(i < j) {
            while(i < j && !isalnum(s[i])) {
                i++;
            }
            while(i < j && !isalnum(s[j])) {
                j--;
            }
            if(i < j && tolower(s[i]) != tolower(s[j])) {
                return false;
            } else {
                i++;
                j--;
            }
        }
        return true;
    }
};

class Solution {

public:

bool isPalindrome(string s) {

if(s.length() == 0)

return true;

int i = 0, j = s.length() - 1;

while(i < j) {

while(i < j && !isalnum(s[i])) {

i++;

}

while(i < j && !isalnum(s[j])) {

j--;

}

if(i < j && tolower(s[i]) != tolower(s[j])) {

return false;

} else {

i++;

j--;

}

return true;

}

};

LeetCode 6. ZigZag Conversion

The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P A H N
A P L S I I G
Y I R
And then read line by line: “PAHNAPLSIIGYIR”
Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);
convert(“PAYPALISHIRING”, 3) should return “PAHNAPLSIIGYIR”.

class Solution {
public:
    string convert(string s, int numRows) {
        string t[numRows];
        string ans;
        int i = 0;
        while(i < s.length()) {
            for(int j = 0; j < numRows && i < s.length(); j++)
                t[j] += s[i++];
            for(int k = numRows - 2; k > 0 && i < s.length(); k--)
                t[k] += s[i++];
        }
        for(int j = 0; j < numRows; j++) {
            ans += t[j];
        }
        return ans;
    }
};

class Solution {

public:

string convert(string s, int numRows) {

string t[numRows];

string ans;

int i = 0;

while(i < s.length()) {

for(int j = 0; j < numRows && i < s.length(); j++)

t[j] += s[i++];

for(int k = numRows - 2; k > 0 && i < s.length(); k--)

t[k] += s[i++];

}

for(int j = 0; j < numRows; j++) {

ans += t[j];

}

return ans;

}

};