前言
本文为leetcode上的题目简单分析，仅作记录，欢迎提出建议，共同学习交流。题目的源代码和测试用例可以在leetcodeWithC下载
001 Two Sum
题目：
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
释义：
给定整型数组，返回两个数的下标，使得这两个数相加得到特定的值。
假设每个给定的数组只能找到一组满足条件的结果，同时，不能使用同一个数两次。
分析：
题大意为，在一组数组中，找到两个数，使得这两个数的和等于特定值，并返回下标。可以从第一个数开始，循环与后面的每一个相加，与结果比较，比较成功则返回。
例如，输入[1,7,11,15],目标值26，那么循环计算1+7,1+11,1+15，7+11,7+15……，直到得到目标值。
代码如下：
int* twoSum(int* nums, int numsSize, int target) {
int loop = 0;
int inloop = 0;
int* result = NULL;
result =(int*) malloc(2*sizeof(int));
memset(result,0,2*sizeof(int));
printf("numsSize=%d\n",numsSize);
if(NULL == nums || numsSize==0)
{
return result;
}
for(loop = 0;loop < numsSize;loop++)
{
for(inloop = loop+1;inloop <numsSize;inloop++)
{
if(*(nums+loop)+*(nums+inloop) == target)
{
if(NULL != result)
{
*result = loop;
*(result+1) = inloop;
}
return result;
}
}
}
return result;
}002 Add Two Numbers
题目：
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
释义：
给定两个非空链表代表两个非负整数，整数的各位数以逆序存储在链表的每个节点中。将这两个数相加，并返回结果链表。
分析
题意较清晰，是将用链表形式的两个整数进行相加，并返回链表结果。
需要注意的主要有以下几点
1.加完之后需要给下一位子进位。
2.如果链表只有一位，直接计算结果，提高效率。
3.考虑两个链表长度不一样的场景
代码如下：
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
unsigned int len = sizeof(struct ListNode);
struct ListNode* head = NULL;
struct ListNode* temp = NULL;
int tempValue = 0;
int flag = 0;
/×个位数相加的情况×/
if(NULL == l1->next||NULL == l2->next)
{
head = (struct ListNode*)malloc(len);
memset(head,0,len);
head->val = (l1->val + l2->val)%10;
/×获取进位×/
flag = (l1->val + l2->val)/10;
temp = head;
l1 = l1->next;
l2 = l2->next;
}
else
{
/×非个位数计算×/
while(NULL != l1&& NULL != l2)
{
tempValue = l2->val+l1->val;
struct ListNode* node = (struct ListNode*)malloc(len);
memset(node,0,len);
node->next = NULL;
node->val = (tempValue+flag)%10;
if(NULL == head)
{
head = node;
temp = head;
}
{
temp->next = node;
temp = temp->next;
}
flag = (tempValue+flag)/10;
l1 = l1->next;
l2 = l2->next;
}
}
/×l1的位数大于l2×/
if(NULL != l1)
{
while(NULL != l1)
{
struct ListNode* node = (struct ListNode*)malloc(len);
memset(node,0,len);
node->next = NULL;
node->val = (l1->val+flag)%10;
temp->next = node;
temp = temp->next;
flag = (l1->val+flag)/10;
l1=l1->next;
}
}
/×l2的位数大于l1×/
else if(NULL != l2)
{
while(NULL != l2)
{
struct ListNode* node = (struct ListNode*)malloc(len);
memset(node,0,len);
node->next = NULL;
node->val = (l2->val+flag)%10;
temp->next = node;
temp = temp->next;
flag =(l2->val+flag)/10;
l2=l2->next;
}
}
else{}
/×如果前面有进位，记得加上×/
if(flag != 0)
{
temp->next = (struct ListNode*)malloc(len);
memset(temp->next,0,len);
temp->next->val = flag;
}
return head;
}258 Add Digits
题目：
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
释义：
计算各位数之和，直到最后只有一位。
分析：
计算各位数之和，采用递归，计算一次后，再次调用，直到结果位各位数。
代码如下：
int addDigits(int num) {
int temp = num;
if(0 == num/10)
{
return num;
}
else
{
num =0;
while(0 != temp/10)
{
num +=temp%10;
temp = temp/10;
}
num = num+temp;
return addDigits(num);
}
}344 Reverse String
题目：
Write a function that takes a string as input and returns the string reversed.
Example:
Given s = “hello”, return “olleh”.
释义：
字符串翻转
分析
头尾对应位置的字符位置调换
代码如下：
char* reverseString(char* s)
{
unsigned int len = strlen((const char*)s);
char result[len+1];
memset(result,0,len);
int loop = 0;
for(loop = len-1; loop >= 0;loop--)
{
*(result+(len-loop-1)) = *(s+loop);
}
result[len] = '\0';
memcpy(s,result,len);
return s;
}463 Hamming Distance
题目：
The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given two integers x and y, calculate the Hamming distance.
Note:
0 ≤ x, y < 231.
Example:
Input: x = 1, y = 4
Output: 2
Explanation:
1 (0 0 0 1)
4 (0 1 0 0)
? ?
The above arrows point to positions where the corresponding bits are different.
释义：
分析：
两个数的汉明距离，可以理解为，二进制的情况下，两个数异或之后的数的1的个数。
比如例子中，1和4,0001与0100异或得：0101，而0101中1的个数，即为汉明距离，可以理解位，从0001，变成0100，需要改变的位数。
代码如下：
““
int hammingDistance(int x, int y) {
int result = 0;
/get the temp result/
int temp = x^y;
/calc the 1 of temp result/
while (temp != 0)
{
if (temp % 2 == 1)
{
result++;
}
temp = temp >>1;
}
return result;
}
### 476 Number Complement
##### 题目：
Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Note:
The given integer is guaranteed to fit within the range of a 32-bit signed integer.
You could assume no leading zero bit in the integer’s binary representation.
Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
##### 释义：
给定一个正整数，输出它的补全整数。补全策略是翻转它的每个比特位，得到补全整数。
##### 分析：
对于补全，以5为例，101，每一位翻转得到，010，即结果为2，那么，010+101 = 111，2的3次方。再以35为例，10 0011，翻转得到011100，即有100011+011100=111111,2的6次方，那么不难得到，其实补全数，就是用2的n次方，减去该数，其中，n为该数二进制表示的位数。
##### 代码如下：
[476 Number Complement](https://github.com/yanbinghu/LeetCodeWithC/blob/master/leetcode/src/476_Number_Complement.c)int findComplement(int num)
{
int temp = num;
int i = 0;
for(i=0;temp!=0;temp/=2,i++)
{}
/×pow，库函数×/
return pow(2,i)-num-1;
}
“`