剑指 Offer第四天

剑指 Offer 27. 二叉树的镜像

/**
 * 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:
    TreeNode* mirrorTree(TreeNode* root) {
        if (!root) return NULL;
        swap(root->left, root->right);
        mirrorTree(root->left);
        mirrorTree(root->right);
        return root;
    }
};

剑指 Offer 28. 对称的二叉树

/**
 * 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:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        return dfs(root->left, root->right);
    }

    bool dfs(TreeNode* l, TreeNode* r) {
        if (!l && !r) return true;
        else if (!l || !r || l->val != r->val) return false;
        return dfs(l->left, r->right) && dfs(l->right, r->left);
    }
};

剑指 Offer 20. 表示数值的字符串

思路：分情况判断

首先将字符串空格删去，如果此时空，那么$returnfalse$，将最前面的$+$和$-$去掉，特判如果只有一个.或者此时是空字符串就$returnfalse$，遍历每一个字符详细的介绍在代码中

class Solution {
public:
    bool isNumber(string s) {
        int i = 0, j = s.size() - 1;
        while (i <= j && s[i] == ' ') i ++ ;
        while (i <= j && s[j] == ' ') j -- ;
        if (i > j) return false; // " "
        s = s.substr(i, j - i + 1);
        if (s[0] == '+' || s[0] == '-') s = s.substr(1); // "+12341234"
        if (s.empty() || s[0] == '.' && s.size() == 1) return false; // "."
        int dot = 0, e = 0;
        for (int i = 0; i < s.size(); i ++ ) {
            if (s[i] >= '0' && s[i] <= '9') ;
            else if (s[i] == '.') {
                dot ++ ;
                if (dot > 1 || e) return false; // ".1234.21" "e123.123"
            }
            else if (s[i] == 'e' || s[i] == 'E') {
                e ++ ;
                if (!i || i + 1 == s.size() || e > 1 || s[i - 1] == '.' && i == 1) return false; // "e" "e1" "ee" ".e"
                if (s[i + 1] == '+' || s[i + 1] == '-') {
                    if (i + 2 == s.size()) return false; // "12424e+"
                    i ++ ; // 过滤+ -
                }
            }
            else return false;
        }
        return true;
    }
};

剑指 Offer 21. 调整数组顺序使奇数位于偶数前面

思路一：快速排序的思想

类似快速排序的思想，从前往后找，找到第一个偶数，从后往前找找到第一个奇数，如果此时$l<r$交换，直到$l>=r$

class Solution {
public:
    vector<int> exchange(vector<int>& nums) {
        int n = nums.size();
        if (!n) return {};
        int l = 0, r = n - 1;
        while (l < r) {
            while (l + 1 < n && (nums[l] & 1)) l ++ ;
            while (r - 1 >= 0 && (nums[r] & 1) == 0) r -- ;
            if (l < r) swap(nums[l], nums[r]);
        }
        return nums;
    }
};

思路二：双指针

class Solution {
public:
    vector<int> exchange(vector<int>& nums) {
        if (nums.size() == 0) return {};
        int l = 0, r = 0;
        while (r < nums.size()) {
            if (nums[r] & 1) {
                swap(nums[l], nums[r]);
                l ++ ;
            }
            r ++ ;
        }
        return nums;
    }
};

剑指 Offer 15. 二进制中1的个数

思路一：lowbit函数

每一次将最后一个一删去

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int res = 0;
        while (n) {
            n -= n & (-n);
            res ++ ;
            cout << n << endl;
        }
        return res;
    }
};

思路二：巧妙使用位运算

class Solution {
public:
    int hammingWeight(uint32_t n) {
        int res = 0;
        while (n) {
            n = n & (n - 1);
            res ++ ;
            cout << n << endl;
        }
        return res;
    }
};