Leetcode--easy系列10

最新推荐文章于 2021-10-19 14:35:06 发布

转载最新推荐文章于 2021-10-19 14:35:06 发布 · 75 阅读

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原文链接：http://www.cnblogs.com/yfceshi/p/7159957.html

本文解析了四道LeetCode经典题目：判断两字符串是否同构、反转单链表、计算两个矩形覆盖的总面积及翻转二叉树。通过代码示例详细介绍了每道题目的解题思路。

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#205 Isomorphic Strings

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

For example,
Given "egg", "add", return true.

Given "foo", "bar", return false.

Given "paper", "title", return true.

Note:
You may assume both s and t have the same length.

推断2个字符串结构是否同样（默认长度相等）。

最開始我是这样想的，将两个同构不同型的字符串按规则变为同构同型的字符串，比較转换后的字符串是否等。

如paper 'p'变为1，‘a’变为2，‘e’变为3 ‘r’变为4.则字符串变为 12134 title 类似变为12134。相等说明同构。

还有一种思路是分别遍历两个字符串，利用hash表中的s[i]位置存储t[i]中的字符。当下一次s字符串中再次出现s[j] ==s[i] 时，对于 t[j] 位置上的字符应该和先前的字符 t[i] 同样。

//0ms
bool isIsomorphic(char* s, char* t) {
    int hash[128] = {0};
    int i;
    for( i = 0; s[i] != '\0'; i++)
    {
        if(!hash[s[i]])
            hash[s[i]] = t[i];
        else if (hash[s[i]] != t[i])
            return false;
    }
    memset(hash,0,sizeof(hash));
    for( i =0; t[i] != '\0'; i++)
    {
        if(!hash[t[i]])
            hash[t[i]] = s[i];
        else if (hash[t[i]] != s[i])
            return false;
    }
    return true;
}

#206 Reverse Linked List

Reverse a singly linked list.

//0ms
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* reverseList(struct ListNode* head) {
    struct ListNode *newhead,*p,*new_p,*r;
	newhead->next = head;
	p = head;
	r = NULL;
	while(p)
	{
		new_p = p->next;
		p->next = r;	
		r = p;
		p = new_p;
	}
	return r;  
}

#223 Rectangle Area

Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Assume that the total area is never beyond the maximum possible value of int.

2个对角顶点能够确定一个长方形，给定4个点的坐标。求它们构成的2个长方形覆盖的面积。

关键在于怎样依据坐标的相对大小来确定2个长方形是否相互覆盖。

//12ms
int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
    int area = (C-A)*(D-B) + (G-E)*(H-F);
    int top,bottom,left,right,cover;
    if(A>=G || C<=E || B>=H || D<=F)
        return area;
    top = (D<=H)?D:H;
    bottom = (B>=F)?B:F;
    left = (A>=E)?

A:E; right = (C<=G)?

C:G; cover = (top - bottom)*(right-left); return area-cover; }

#226 Invert Binary Tree

Invert a binary tree.

     4
   /   \
  2     7
 / \   / \
1   3 6   9

     4
   /   \
  7     2
 / \   / \
9   6 3   1

//0ms
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* invertTree(struct TreeNode* root) {
    struct TreeNode* p;
    if(!root)
        return NULL;
    else if(!root->left && !root->right)    
        return root;
     p = root->left;
     root->left = root->right;
     root->right = p;
     
     invertTree(root->left);
     invertTree(root->right);
     return root;
}

转载于:https://www.cnblogs.com/yfceshi/p/7159957.html