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;
}

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;  
}

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.

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;    
}

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;
}