Leetcode--easy系列10
#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.
最開始我是這樣想的，將兩個同構不同型的字符串按規則變為同構同型的字符串。比較轉換後的字符串是否等。
如paper ‘p‘變為1，‘a’變為2，‘e’變為3 ‘r’變為4.則字符串變為 12134 title 類似變為12134。相等說明同構。
還有一種思路是分別遍歷兩個字符串，利用hash表中的s[i]位置存儲t[i]中的字符，當下一次s字符串中再次出現s[j] ==s[i] 時，對於 t[j] 位置上的字符應該和先前的字符 t[i] 同樣。
//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個對角頂點能夠確定一個長方形，給定4個點的坐標。求它們構成的2個長方形覆蓋的面積。
關鍵在於怎樣依據坐標的相對大小來確定2個長方形是否相互覆蓋。
//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 9to
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;
}
Leetcode--easy系列10