leetcode简单题目两道(4)

心情还是有问题，保持每日更新，只能如此了。

Problem

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

Example:

Given binary tree {,,,#,#,,},

····
   / \

    /  \

Result

return its level order traversal as:

[
  [],
  [,],
  [,]
]
Code

/**
 * 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:
    int HeightOfTree(TreeNode* root) {
        int result = ;
        if (root == NULL) {
            return result;
        }
        int left = HeightOfTree(root->left) + ;
        int right = HeightOfTree(root->right) + ;
        result = left > right ? left : right;
        return result;
    }
    void calLevelOrder(TreeNode* root, int level, vector<vector<int> >& result) {
        if (root == NULL) {
            return;
        }
        result[level].push_back(root->val);
        calLevelOrder(root->left, level + , result);
        calLevelOrder(root->right, level + , result);
    }
    vector<vector<int> > levelOrder(TreeNode* root) {
        vector<vector<int> > result;
        vector<int> v;
        if (root == NULL) {
            return result;
        }
        int height = HeightOfTree(root);
        for (int i = ; i < height; i++) {
            result.push_back(v);
        }
        calLevelOrder(root, , result);
        return result;
    }
};

Problem1

Write a function that takes a string as input and returns the string reversed.

Example:

Given s = "hello", return "olleh".

Code

class Solution {
public:
    string reverseString(string s) {
        if (s.size() <= ) {
            return s;
        }
        int i = , j = s.size() - ;
        char ch;
        while (i < j) {
            ch = s[i];
            s[i] = s[j];
            s[j] = ch;
            i++;
            j--;
        }
        return s;
    }
};
Problem2

Write a function that takes a string as input and reverse only the vowels of a string.

Example :

Given s = "hello", return "holle".

Example :

Given s = "leetcode", return "leotcede".

Code

class Solution {
public:
    string reverseVowels(string s) {
        if (s.size() <= ) {
            return s;
        }
        int i = , j = s.size() - ;
        char ch;
        while (i < j) {
            while(i < j && s[i] != &#;a&#; && s[i] != &#;e&#; && s[i] != &#;i&#; && s[i] != &#;o&#; && s[i] != &#;u&#;
            && s[i] != &#;A&#; && s[i] != &#;E&#; && s[i] != &#;I&#; && s[i] != &#;O&#; && s[i] != &#;U&#;) {
                i++;
            }
            while(i < j && s[j] != &#;a&#; && s[j] != &#;e&#; && s[j] != &#;i&#; && s[j] != &#;o&#; && s[j] != &#;u&#;
            && s[j] != &#;A&#; && s[j] != &#;E&#; && s[j] != &#;I&#; && s[j] != &#;O&#; && s[j] != &#;U&#;) {
                j--;
            }
            ch = s[i];
            s[i] = s[j];
            s[j] = ch;
            i++;
            j--;
        }
        return s;
    }
};
说明

与上面同样的逻辑，只是出现特定的字符才交换位置

PS:存货快用完了。。。。