66. Plus One
description：
Given a non-empty array of digits representing a non-negative integer, plus one to the integer.
The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.
You may assume the integer does not contain any leading zero, except the number 0 itself.
思路：
由低到高位分别进行求和和溢出操作，求出最后的数字，若最后一位溢出，则特殊处理，否则直接拷贝返回。
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int* plusOne(int* digits, int digitsSize, int* returnSize) {
int i;
int pre = 1;
int temp;
int *p = NULL;
for(i = digitsSize -1; i >= 0; i--){
temp = (digits[i]+pre)%10;
pre = (digits[i] + pre) / 10;
digits[i] = temp;
}
if(pre){
*returnSize = digitsSize+1;
p = malloc(sizeof(int) * (*returnSize));
if(p){
memset(p, 0, *returnSize);
p[0] = 1;
return p;
}
}
p = malloc(sizeof(int) * digitsSize);
if(p){
memcpy(p, digits, sizeof(int) * digitsSize);
}
*returnSize = digitsSize;
return p;
}
441. Arranging Coins
description：
You have a total of n coins that you want to form in a staircase shape, where every k-th row must have exactly k coins.
Given n, find the total number of full staircase rows that can be formed.
n is a non-negative integer and fits within the range of a 32-bit signed integer.
思路：等差数列，注意乘法溢出。
int arrangeCoins(int n) {
long long i = 0;
while(((i*i + i)>>1) <= n){
i++;
}
return i-1;
}
414. Third Maximum Number
description：
Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).
思路：
插入法，分类讨论。
int thirdMax(int* nums, int numsSize) {
int max_num[3] = {};
int num = 0;
int i;
for(i = 0; i < numsSize; i++){
if(0 == num){
max_num[0] = nums[i];
num++;
}else if(1 == num){
if(nums[i] > max_num[0]){
max_num[1] = nums[i];
num++;
}else if(nums[i] < max_num[0]){
max_num[1] = max_num[0];
max_num[0] = nums[i];
num++;
}
}else if(2 == num){
if(nums[i] > max_num[1]){
max_num[2] = nums[i];
num++;
}else if(nums[i] > max_num[0] && nums[i] != max_num[1]){
max_num[2] = max_num[1];
max_num[1] = nums[i];
num++;
}else if(nums[i] < max_num[0]){
max_num[2] = max_num[1];
max_num[1] = max_num[0];
max_num[0] = nums[i];
num++;
}
}else{
if(max_num[2] < nums[i]){
max_num[0] = max_num[1];
max_num[1] = max_num[2];
max_num[2] = nums[i];
}else if(nums[i] > max_num[1] && nums[i] != max_num[2]){
max_num[0] = max_num[1];
max_num[1] = nums[i];
}else if(nums[i] > max_num[0] && nums[i] != max_num[1] && nums[i] != max_num[2]){
max_num[0] = nums[i];
}
}
}
return (num == 3) ? max_num[0] : max_num[num-1];
}
349. Intersection of Two Arrays
description：
Given two arrays, write a function to compute their intersection.
思路：
快速排序，然后依次对比。只对最小值进行了优化。
最佳思路：
依旧是快速排序，不过是动态进行对比，跳过不需要进行对比的部分，直接申请内存。
/**
* Return an array of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
/**
int compare(const void * a, const void * b) {
return ( *(int*)a - *(int*)b );
}
int *mem_set(int *p, int val, int *num){
int *p_temp;
if(NULL == p){
p = malloc(sizeof(int));
if(p){
*p = val;
*num = 1;
}else{
return NULL;
}
return p;
}else{
p_temp = malloc(sizeof(int) * (*num+1));
if(p_temp){
memcpy(p_temp, p, sizeof(int) * (*num));
p_temp[*num] = val;
(*num)++;
}else{
free(p);
return NULL;
}
return p_temp;
}
}
int* intersection(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize) {
if(NULL == nums1 || 0 == nums1Size || NULL == nums2 || 0 == nums2Size){
goto label;
}
int pre;
int i, j;
int *p = NULL;
qsort(nums1, nums1Size, sizeof(int), compare);
qsort(nums2, nums2Size, sizeof(int), compare);
pre = nums1[0] - 1;
*returnSize = 0;
for(i = 0; i < nums1Size; i++){
if(nums1[i] < nums2[0]){
pre = nums1[i];
continue;
}
if(pre == nums1[i]){
continue;
}
for(j = 0; j < nums2Size; j++){
if(nums1[i] == nums2[j]){
p = mem_set(p, nums1[i], returnSize);
if(NULL == p){
goto label;
}
break;
}
}
pre = nums1[i];
}
return p;
label:
*returnSize = 0;
return NULL;
}
*/
int compare(const void *a,const void *b){
return *(int *)a-*(int *)b;
}
int* intersection(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize) {
int i,j,size,*Rarray;
size = nums1Size < nums2Size? nums1Size: nums2Size;
Rarray=(int *)malloc(size*sizeof(int));
*returnSize=0;
qsort(nums1,nums1Size,sizeof(int),compare);
qsort(nums2,nums2Size,sizeof(int),compare);
for(i=0,j=0;i<nums1Size&&j<nums2Size; ){
while(i<nums1Size&&nums1[i]<nums2[j])
i++;
if(nums1[i] == nums2[j]){
Rarray[*returnSize]=nums1[i];
while(i<nums1Size&&nums1[i]==Rarray[*returnSize])
i++;
while(j<nums2Size&&nums2[j]==Rarray[*returnSize])
j++;
(*returnSize)++;
}
while(j<nums2Size&&nums1[i]>nums2[j])
j++;
}
return Rarray;}
674. Longest Continuous Increasing Subsequence
description：
Given an unsorted array of integers, find the length of longest continuous increasing subsequence (subarray).
思路：
前后对比，更新最大值。
int findLengthOfLCIS(int* nums, int numsSize) {
if(NULL == nums || 0 == numsSize){
return 0;
}
int cur_subsequence = 1;
int max_subsequence = 1;
int i;
for(i = 0; i < numsSize-1; i++){
if(nums[i] < nums[i+1]){
cur_subsequence++;
}else{
if(cur_subsequence > max_subsequence){
max_subsequence = cur_subsequence;
}
cur_subsequence = 1;
}
}
if(cur_subsequence > max_subsequence){
max_subsequence = cur_subsequence;
}
return max_subsequence;
}
459. Repeated Substring Pattern
description：
Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.
思路：
假设是重复的，那么先找出最大重复字符串，从后往前对比，最长不超过字符串的一半，然后找到最大子串后，对比中间部分是否符合。
最佳思路：
假设重复，一份一份的对比。比如字符串长为6，则分别测试其整除数是否符合重复条件，即1,2,3.
/*bool repeatedSubstringPattern(char* s) {
int i, k;
char substr[10000] = {};
int substr_len_cur = 0;
int substr_len_max = 0;
int str_len = strlen(s);
int j = str_len / 2;
for(k = 0; k < j; k++){
for(i = 0; i < k+1; i++){
if(s[i] == s[str_len-1-k+i]){
substr[i] = s[i];
substr_len_cur++;
}else{
substr_len_cur = substr_len_max;
break;
}
}
substr_len_max = substr_len_cur;
substr_len_cur = 0;
}
for(i = substr_len_max, k = 0; i < str_len-substr_len_max; i++, k++){
if(substr[k] != s[i]){
return false;
}
}
return true;
}*/
bool repeatedSubstringPattern(char* s) {
int size = strlen(s);
int found = true, start = 0;
int i = 0;
if(size <= 1)
return false;
for(i=1; i <= size / 2; i++)
{
if(size % i == 0)
{
found = true;
for(start = i; start + i <= size; start += i)
{
if(memcmp(s, s + start, i) != 0)
{
found = false;
break;
}
}
if(found)
{
return true;
}
}
}
return false;
}
231. Power of Two
description：
Given an integer, write a function to determine if it is a power of two.
思路：
确定某位是否被set，然后确认其他位没有被set。位的基本操作。
#define TEST_THE_BIT(x, n) x & (1 << n)
#define TEST_OTHER_BIT(x, n) x & ~(1 << n)
bool isPowerOfTwo(int n) {
if(n < 1){
return false;
}
int i;
for(i = 0; i< 31; i++){
if(TEST_THE_BIT(n, i)){
if(TEST_OTHER_BIT(n, i)){
return false;
}else{
return true;
}
}
}
return false;
}