目录
第1题：单词规律
第2题：找不同
第3题：在排序数组中查找元素的第一个和最后一个位置
第4题：使用最小花费爬楼梯
第5题：寻找峰值
第6题：字符串中的第一个唯一字符
第7题：两个数组的交集 II
第8题：分发饼干
第9题：旋转图像
第10题：矩阵置零
力扣(LeetCode)定期刷题，每期10道题，业务繁重的同志可以看看我分享的思路，不是最高效解决方案，只求互相提升。
试题要求如下：
bool wordPattern(char * pattern, char * str){
char **hash = (char **)malloc(26 * sizeof(char*));
for (int i = 0; i < 26; ++i)
{
hash[i] = (char*)malloc(64 * sizeof(char));
memset(hash[i], 0, 64 * sizeof(char));
}
int len = strlen(pattern);
for (int i = 0; i < len; ++i)
{
char *p = str;
while (p && *p != 0 && *p != ' ') ++p;
if (' ' == *p) *p++ = 0;
if (strlen(str) == 0)
return false;
int pos = pattern[i] - 'a';
if (strlen(hash[pos]) == 0)
{
for (int j = 0; j < 26; ++j)
{
if (j != pos && strlen(hash[j]) > 0)
{
if (strcmp(hash[j], str) == 0)
return false;
}
}
strcpy(hash[pos], str);
}
else
{
if (strcmp(hash[pos], str) != 0)
return false;
}
str = p;
}
if (strlen(str) > 0)
return false;
return true;
} 运行效率如下所示：
解题思路：
char findTheDifference(char* s, char* t) {
int n = strlen(s), m = strlen(t);
int as = 0, at = 0;
for (int i = 0; i < n; i++) {
as += s[i];
}
for (int i = 0; i < m; i++) {
at += t[i];
}
return at - as;
} int* searchRange(int* nums, int numsSize, int target, int* returnSize){
int *ret=(int *)malloc(sizeof(int)*2);
ret[0]=-1;
ret[1]=-1;
*returnSize=2;
if(numsSize==0||target<nums[0]||target>nums[numsSize-1]){
return ret;
}
int left=0, right=numsSize-1, mid=0, head=0, tail=0;
while(left<=right){
mid=(left+right)/2;
if(nums[mid]>=target){
right=mid-1;
}
else{
left=mid+1; //mid存的就是第一个>=target的那个值的索引
}
}
head=left;
left=0, right=numsSize-1, mid=0;
while(left<=right){
mid=(left+right)/2;
if(nums[mid]<=target){
left=mid+1;
}
else{
right=mid-1;
}
}
tail=right;
if(nums[head]==target){ //存在这个数
ret[0]=head;
ret[1]=tail;
}
return ret;
} int minCostClimbingStairs(int* cost, int costSize) {
int prev = 0, curr = 0;
for (int i = 2; i <= costSize; i++) {
int next = fmin(curr + cost[i - 1], prev + cost[i - 2]);
prev = curr;
curr = next;
}
return curr;
} int findPeakElement(int *nums, int numsSize)
{
if (!nums || numsSize < 1) {
return 0;
}
if (numsSize == 1) {
return 0;
}
if (nums[0] > nums[1]) {
return 0;
}
if (nums[numsSize - 1] > nums[numsSize - 2]) {
return numsSize - 1;
}
for (int i = 1; i < numsSize - 1; i++) {
if (nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) {
return i;
}
}
return -1;
} 两次遍历，一次记录字符出现次数，一次找出第一个出现一次的索引。
int firstUniqChar(char * s){
if(s == NULL || strlen(s) == 0) return -1;
int len = strlen(s);
if(len == 1) return 0;
int recode[26] = {0};
//记录字符出现次数
for(int i=0;i<len;i++){
recode[s[i]-'a']++;
}
//找出第一个出现一次的索引
for(int i=0;i<len;i++){
if(recode[s[i]-'a'] == 1){
return i;
}
}
return -1;
} /**
* Note: The returned array must be malloced, assume caller calls free().
*/
struct cell
{
int value;
int times1;
int times2;
};
int* intersect(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize){
int i = 0, cur, mapSize = nums1Size + nums2Size, *ret = (int *)malloc(sizeof(int) * (nums1Size + nums2Size));
*returnSize = 0;
struct cell *hashMap = (struct cell *)memset(malloc(sizeof(struct cell) * mapSize), 0, sizeof(struct cell) * mapSize);
for (i = 0; i < nums1Size; i++)
{
cur = (nums1[i] > 0 ? 1 : -1) * (nums1[i] % mapSize);
while((hashMap[cur].times1 || hashMap[cur].times2) && hashMap[cur].value != nums1[i])
{
cur++;
cur = cur == mapSize ? 0 : cur;
}
if (hashMap[cur].times1 == 0 && hashMap[cur].times2 == 0)
hashMap[cur].value = nums1[i];
hashMap[cur].times1++;
}
for (i = 0; i < nums2Size; i++)
{
cur = (nums2[i] > 0 ? 1 : -1) * (nums2[i] % mapSize);
while((hashMap[cur].times1 || hashMap[cur].times2) && hashMap[cur].value != nums2[i])
{
cur++;
cur = cur == mapSize ? 0 : cur;
}
if (hashMap[cur].times1 == 0 && hashMap[cur].times2 == 0)
hashMap[cur].value = nums2[i];
hashMap[cur].times2++;
}
for (i = 0; i < mapSize; i++)
while ((hashMap[i].times2--) && (hashMap[i].times1--))
ret[(*returnSize)++] = hashMap[i].value;
return ret;
} int compare(const void * a, const void * b)
{
return ( *(int*)b - *(int*)a );
}
int findContentChildren(int* g, int gSize, int* s, int sSize){
int count=0;
qsort (g, gSize, sizeof(int), compare);
qsort (s, sSize, sizeof(int), compare);
for(int i = 0, j = 0; i < gSize && j < sSize; i++, j++)
{
if(s[j] >= g[i])
count++;
else
j--;
}
return count;
} void rotate(int** matrix, int matrixSize, int* matrixColSize) {
int matrix_new[matrixSize][matrixSize];
for (int i = 0; i < matrixSize; i++) {
for (int j = 0; j < matrixSize; j++) {
matrix_new[i][j] = matrix[i][j];
}
}
for (int i = 0; i < matrixSize; ++i) {
for (int j = 0; j < matrixSize; ++j) {
matrix[j][matrixSize - i - 1] = matrix_new[i][j];
}
}
} void setZeroes(int** matrix, int matrixSize, int* matrixColSize){
int i = 0;
int j = 0;
int iRow = matrixSize;
int iCol = matrixColSize[0];
int row[iRow];
int col[iCol];
memset(row, 0x00, sizeof(int) * iRow);
memset(col, 0x00, sizeof(int) * iCol);
//1,遍历一遍，找出所有为0的元素，并在行，列数组中标记
for (i = 0; i < iRow; i++)
{
for (j = 0; j < iCol; j++)
{
if (matrix[i][j] == 0)
{
row[i] = 1;
col[j] = 1;
}
}
}
//2,遍历数组，按照行，列中的结果修改矩阵
for (i = 0; i < iRow; i++)
{
for (j = 0; j < iCol; j++)
{
if (row[i] == 1)
{
matrix[i][j] = 0;
}
if (col[j] == 1)
{
matrix[i][j] = 0;
}
}
}
}