柳婼の blog – 第243页 – 我不管，反正我最萌～

presentViewController:navigationController animated:YES completion:^(void)

xcode 6.0新增：

- (void)presentModalViewController:(UIViewController *)modalViewController animated:(BOOL)animated;

- (void)dismissModalViewControllerAnimated:(BOOL)animated;

- (void)presentModalViewController:(UIViewController *)modalViewController animated:(BOOL)animated;

- (void)dismissModalViewControllerAnimated:(BOOL)animated;

取而代之的是：

[self presentViewController:navigationController
                       animated:YES
                     completion:^(void){
                         // Code
                     }];

[self dismissViewControllerAnimated:YES
                             completion:^(void){
                                 // Code
                             }];

[self presentViewController:navigationController

animated:YES

completion:^(void){

// Code

}];

[self dismissViewControllerAnimated:YES

completion:^(void){

// Code

}];

新接口的差别是提供了一个参数，允许你传入一个block。这个block的回调方法在VC的viewWillDisappear方法后调用。也就是被隐藏的VC对象被释放后运行回调。

这样做的好处：可以方便做多个UI效果之间的衔接和转换。无需的时候可以写 nil。

LeetCode 144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> s;
        vector<int> v;
        if(root == NULL) {
            return v;
        }
        TreeNode *p = root;
        s.push(p);
        while(!s.empty()) {
            p = s.top();
            s.pop();
            v.push_back(p->val);
            if(p->right != NULL) {
                s.push(p->right);
            }
            if(p->left != NULL) {
                s.push(p->left);
            }
        } 
        return v;
    }
};

/**

* Definition for a binary tree node.

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

* TreeNode(int x) : val(x), left(NULL), right(NULL) {}

* };

class Solution {

public:

vector<int> preorderTraversal(TreeNode* root) {

stack<TreeNode*> s;

vector<int> v;

if(root == NULL) {

return v;

}

TreeNode *p = root;

s.push(p);

while(!s.empty()) {

p = s.top();

s.pop();

v.push_back(p->val);

if(p->right != NULL) {

s.push(p->right);

}

if(p->left != NULL) {

s.push(p->left);

}

return v;

}

};

LeetCode319. Bulb Switcher

319. Bulb Switcher
My Submissions QuestionEditorial Solution
Total Accepted: 17314 Total Submissions: 42815 Difficulty: Medium
There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it’s off or turning off if it’s on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Example:

Given n = 3.

At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off].

So you should return 1, because there is only one bulb is on.

分析：可知题意是想判断1~n中那些数的因子个数是奇数，一个数 a = b x c 当 b = c 的时候 a 的因子个数就是奇数，所以就是判断1~n当中有多少个是某数的平方～

class Solution {
public:
    int bulbSwitch(int n) {
        int ans = 0;
        for(int i = 1; i * i <= n; i++) {
            ans++;
        }
        return ans;
    }
};

class Solution {

public:

int bulbSwitch(int n) {

int ans = 0;

for(int i = 1; i * i <= n; i++) {

ans++;

}

return ans;

}

};

LeetCode 8. String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

spoilers alert… click to show requirements for atoi.

Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

class Solution {
public:
    int myAtoi(string str) {
        if(str.length() == 0)
            return 0;
        int i = 0, flag = 1;
        while(isspace(str[i])) {
            i++;
        }
        if(str[i] == '+' || str[i] == '-') {
            if(str[i] == '-')
                flag = 0;
            i++;
        }
        str = str.substr(i);
        for(int j = 0; j < str.length(); j++) {
            if(!isdigit(str[j])) {
                str = str.substr(0, j);
                break;
            }
        }
        if(str.length() == 0)
            return 0;
        if(str.length() > 10) {
            if(flag == 0)
                return -2147483648;
            else
                return 2147483647;
        }
        long long int ans = stoll(str);
        if(flag == 0)
            ans = 0 - ans;
        if(ans > 2147483647) {
            return 2147483647;
        } else if (ans < -2147483648) {
            return -2147483648;
        } else {
            return (int)ans;
        }
    }
};

class Solution {

public:

int myAtoi(string str) {

if(str.length() == 0)

return 0;

int i = 0, flag = 1;

while(isspace(str[i])) {

i++;

}

if(str[i] == '+' || str[i] == '-') {

if(str[i] == '-')

flag = 0;

i++;

}

str = str.substr(i);

for(int j = 0; j < str.length(); j++) {

if(!isdigit(str[j])) {

str = str.substr(0, j);

break;

}

if(str.length() == 0)

return 0;

if(str.length() > 10) {

if(flag == 0)

return -2147483648;

else

return 2147483647;

}

long long int ans = stoll(str);

if(flag == 0)

ans = 0 - ans;

if(ans > 2147483647) {

return 2147483647;

} else if (ans < -2147483648) {

return -2147483648;

} else {

return (int)ans;

}

};

LeetCode165. Compare Version Numbers

Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not “two and a half” or “half way to version three”, it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

class Solution {
public:
    int compareVersion(string version1, string version2) {
        while(!version1.empty() || !version2.empty()) {
            int ver1 = version1.empty() ? 0 : stoi(version1);
            int ver2 = version2.empty() ? 0 : stoi(version2);
            if(ver1 > ver2) {
                return 1;
            } else if(ver1 < ver2) {
                return -1;
            } else {
                int pos1 = version1.find('.');
                int pos2 = version2.find('.');
                version1 = (pos1 == string::npos) ? "" : version1.substr(pos1 + 1);
                version2 = (pos2 == string::npos) ? "" : version2.substr(pos2 + 1);
            }
        }
        return 0;
    }
};

class Solution {

public:

int compareVersion(string version1, string version2) {

while(!version1.empty() || !version2.empty()) {

int ver1 = version1.empty() ? 0 : stoi(version1);

int ver2 = version2.empty() ? 0 : stoi(version2);

if(ver1 > ver2) {

return 1;

} else if(ver1 < ver2) {

return -1;

} else {

int pos1 = version1.find('.');

int pos2 = version2.find('.');

version1 = (pos1 == string::npos) ? "" : version1.substr(pos1 + 1);

version2 = (pos2 == string::npos) ? "" : version2.substr(pos2 + 1);

}

return 0;

}

};

LeetCode 168. Excel Sheet Column Title

Given a positive integer, return its corresponding column title as appear in an Excel sheet.

For example:

1 -> A
2 -> B
3 -> C
…
26 -> Z
27 -> AA
28 -> AB

/*
把 A~Z 对应 0 ~ 25
满 26 进 1
只要看当前 n-1 的值是否 /26 != 0
如果是的 则 将 (n-1) % 26 + 'A' 存储在 s 之前
如果不是 则 退出 while 循环
*/
class Solution {
public:
    string convertToTitle(int n) {
        string s;
        while(n) {
            s = (char)((n - 1) % 26 + 'A') + s;
            n = (n - 1) / 26;
        }
        return s;
    }
};

把 A~Z 对应 0 ~ 25

满 26 进 1

只要看当前 n-1 的值是否 /26 != 0

如果是的则将 (n-1) % 26 + 'A' 存储在 s 之前

如果不是则退出 while 循环

class Solution {

public:

string convertToTitle(int n) {

string s;

while(n) {

s = (char)((n - 1) % 26 + 'A') + s;

n = (n - 1) / 26;

}

return s;

}

};