题目代码区
剑指12
// https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/
// 剑指12，同主站79
// https://leetcode-cn.com/problems/word-search/
// 2021-01-12 22:48
bool exist(char** board, int boardSize, int* boardColSize, char * word){
int wordlength = 0;
while(word[wordlength] != '\0'){
wordlength++;
}
// 设好了wordlength，接下来设一个坐标表
bool found[wordlength];
int co1[wordlength];
int co2[wordlength];
int flag = 0;
for(int i = 0; i <wordlength; i++){
co1[i] = -1;
co2[i] = -1;
found[i] = false;
}
// 接下来设一个与board同大小的布尔表，用来表示board同位置的字母是否被用过
bool* boolboard[boardSize];
for(int x = 0; x <boardSize; x++){
boolboard[x] = (bool *)malloc(sizeof(bool) * boardColSize[x]);
}//为每个小表申请空间
for(int x = 0; x <boardSize; x++){
for(int y =0; y <boardColSize[x];y++){
boolboard[x][y] = false;
}
}//将所有表都设为false
void filltrue(int x, int y){
boolboard[x][y] = true;
}
void fillfalse(int x, int y){
boolboard[x][y] = false;
}
bool checkbool(int x, int y){
return boolboard[x][y];
}
void fillco(int x, int y){
if (flag==wordlength)
return;
co1[flag] = x;
co2[flag] = y;
found[flag] = true;
flag++;
filltrue(x,y);
}
void deleteco(){
flag--;
found[flag] = false;
fillfalse(co1[flag],co2[flag]);
co1[flag] = -1;
co2[flag] = -1;
}
bool check(int now, int x,int y){
return (board[x][y] == word[now]);
}
void initboard(){
while(flag!=0){
deleteco();
}
}
// 套娃函数
// 为的是在x,y周围找到index为now的字母
void taowa(int now, int x, int y){
// x+1阶段
if (flag == wordlength)
return;
while (flag>now){
deleteco();
}
if (x+1 < boardSize){
if( (check(now,x+1,y))&& !checkbool(x+1,y) ){
fillco(x+1,y);
taowa(now+1,x+1,y);
}
}
// x-1阶段
if (flag == wordlength)
return;
while (flag>now){
deleteco();
}
if (x-1 >=0){
if( (check(now,x-1,y))&& !checkbool(x-1,y) ){
fillco(x-1,y);
taowa(now+1,x-1,y);
}
}
// y+1阶段
if (flag == wordlength)
return;
while (flag>now){
deleteco();
}
if(y+1 <boardColSize[x]){
if( (check(now,x,y+1)) && !checkbool(x,y+1)){
fillco(x,y+1);
taowa(now+1,x,y+1);
}
}
// y-1阶段
if (flag == wordlength)
return;
while (flag>now){
deleteco();
}
if(y-1 >=0 ){
if( (check(now,x,y-1)) && !checkbool(x,y-1)){
fillco(x,y-1);
taowa(now+1,x,y-1);
}
}
}
// 流程分为两步
// 1. 找到第一个字母
// 2. 开始套娃找下一个字母
for(int x = 0; x <boardSize; x++){
for(int y =0; y <boardColSize[x];y++){
initboard();
if (check(0,x,y)){
fillco(x,y);
taowa(1,x,y);
}
if(flag == wordlength)
return true;
}
}
return false;
}
139
// 20210113
// https://leetcode-cn.com/problems/word-break/
bool wordBreak(char * s, char ** wordDict, int wordDictSize){
// 求长的函数
int getlen(char *a){
int m = 0;
while(a[m] != '\0')
m++;
return m;
}
// 定义总长
int n = getlen(s);
// 定义一个布尔数组并初始化
bool pd[n+1];
for(int i=1; i<n+1; i++){
pd[i] = false;
}
pd[0] = true;
// 搞个判定函数
bool check(char *a, char *b, int len,int origin){
int i =0;
while(i< len){
if(a[i] != b[i+origin]){
return false;
}else{
i++;
}
}
return true;
}
// 走流程开始
for(int i=0; i<n; i++){
for(int j= i+1; j< n+1; j++){
if( !pd[i]) {
break;
} // i不行的话直接out
if(pd[j]){
continue;
} // j行的话直接out
bool ok = false;
// 接下来轮一圈比较
for (int k= 0; k<wordDictSize; k++){
if(ok)
break;
char* checking = wordDict[k];
int wordlen = getlen(checking);
if(wordlen != j-i)
continue;
ok = check(checking, s, wordlen,i);
}
if(ok)
pd[j] = true;
}
}
return pd[n];
}
98
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool isValidBST(struct TreeNode* root){
// 得到最左边的数
int getsmallest(struct TreeNode* root){
while(root->left != NULL)
root = root->left;
return root->val;
}
int pre = getsmallest(root);
bool first = false;
bool answer = true;
// 写一个中序遍历的函数
void mid(struct TreeNode* root){
if (answer == false)
return;
if(root == NULL)
return;
mid(root->left);
if (first){
if (pre < root->val){
pre = root->val;
}else{
answer = false;
}
}else{
first =true;
}
mid(root->right);
}
mid(root);
return answer;
}
20
// https://leetcode-cn.com/problems/valid-parentheses/
bool isValid(char * s){
bool answer =true;
// 求长的函数
int getlen(char *a){
int m = 0;
while(a[m] != '\0')
m++;
return m;
}
int len= getlen(s);
int stacklen= len/2;
if(stacklen==0)
return false;
int zhan[stacklen];
for (int i = 0; i <stacklen; i++)
zhan[i] =0; // 初试全设为0
int flag = 0;
// 入栈
void push(int a){
if(flag == stacklen){
answer = false;
return;
}
zhan[flag] = a;
flag++;
}
int pop(void){
if(flag == 0){
answer =false;
return;
}
flag--;
int m = zhan[flag];
zhan[flag] = 0;
return m;
}
// 整个过程可以看做一次循环，单次循环里进行入栈或出栈
for (int i = 0; i <len; i++){
if(!answer)
return answer;
if(s[i] == '(')
push(1);
if(s[i] == '[')
push(2);
if(s[i] == '{')
push(3);
if(s[i] == ')')
if(pop() != 1)
answer =false;
if(s[i] == ']')
if(pop() != 2)
answer =false;
if(s[i] == '}')
if(pop() != 3)
answer =false;
}
if(flag == 0)
return answer;
else
return false;
}
62
// https://leetcode-cn.com/problems/unique-paths/
int uniquePaths(int m, int n){
m = m -1;
n = n-1;
if(m == 0 )
return 1;
if(n == 0 )
return 1;
int big, small;
if (m >n ){
big = m, small = n;
}else{
big = n, small = m;
}
long answer=1;
for (int i = big+1; i<=big +small; i++){
answer *=i;
}
for(int i =1; i <= small; i++){
answer /= i;
}
return answer;
}
1
// https://leetcode-cn.com/problems/two-sum/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target, int* returnSize){
for(int i = 0; i< numsSize-1; i++){
for( int j= i+1; j<numsSize;j++){
if( nums[i] + nums[j] ==target){
returnSize[0]=2;
int *m = (int *)malloc(sizeof(int) * 2);
m[0] = i;
m[1] = j;
return m;
}
}
}
return nums;
}
96
// https://leetcode-cn.com/problems/unique-binary-search-trees/
int numTrees(int n){
if(n==1)
return 1;
if(n==2)
return 2;
int p[n+1];
p[0] = 1;
p[1] = 1;
p[2] = 2;
for(int i = 3; i < n +1; i++){
p[i] =0;
for(int j = 0; j<i; j++){
p[i] += (p[j]*p[i-1-j]);
}
}
return p[n];
}
42
// https://leetcode-cn.com/problems/trapping-rain-water/submissions/
// 递归算法，每次求出中间一部分
int trap(int* height, int heightSize){
if (heightSize <= 2)
return 0;
// 开头跟结尾必0
int water[heightSize];
// 把所有水记录为0
for(int i = 0; i <heightSize; i++){
water[i] = 0;
}
int big1,big2;
// 找到最大两个, j必须大于i，否则不好弄
void find2big(int i, int j){
big1 = i;
for(int m = i; m<=j; m++){
if(height[m]> height[big1]){
big1 = m;
}
}// 确定了big1
if(big1 == i )
big2 = i+1;
else
big2 = i;
for(int m = i; m<=j; m++){
if(m==big1)
continue;
if(height[m]> height[big2]){
big2 = m;
}
}// 确定了big2
if(big1>big2){
int temp = big1;
big1 = big2;
big2 = temp;
}
}
// 用于每次求中间一部分的函数，先找到区间内最高的两个柱子
void DOmid(int i, int j){
if((i==j)||((i+1)==j))
return;
find2big(i,j);
int small;
if(height[big1]>height[big2])
small = height[big2];
else
small = height[big1];
for(int k = big1+1; k< big2;k++)
water[k] = small-height[k];
DOmid(i,big1);
DOmid(big2,j);
}
DOmid(0,heightSize-1);
//求和
int sum = 0;
for(int i = 1; i <heightSize-1; i++)
sum += water[i];
return sum;
}
3，剑指48
// language c
// 剑指48
// https://leetcode-cn.com/problems/zui-chang-bu-han-zhong-fu-zi-fu-de-zi-zi-fu-chuan-lcof/
// 同主站3
// https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/
// 滑动窗口
int lengthOfLongestSubstring(char * s){
int count =0;
int max = 0;
int ima = 0;
int start = 0;
// 判断函数
int check(int p, int q){// p 是开始点，q是待判定点
while(p<q){
if(s[p] == s[q])
return p; // 重复
p++;
}
return -1; // 不重复
}
int flag;
while(s[count] != '\0'){
flag = check(start, count);
if(flag==-1){
count++;
}else{ // 结算
ima = count-start;
if( ima >max)
max = ima;
start = flag+1;
count++;
}
}
// 最终结算
ima = count-start;
if( ima >max)
max = ima;
return max;
}
4
// https://leetcode-cn.com/problems/median-of-two-sorted-arrays/
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size){
int index1,index2;
if( (nums1Size+nums2Size)%2 ==1){
index1 = (nums1Size+nums2Size+1)/2;
index2 = index1;
}else{
index1 = (nums1Size+nums2Size)/2;
index2 = index1+1;
}
double an1,an2;
int for1=0,for2=0,count=0,num;
while(count != index1){
if((for1<nums1Size) && (for2<nums2Size)){ // 情况1
if(nums1[for1]<nums2[for2]){
count++;
num = nums1[for1++];
}else{
count++;
num = nums2[for2++];
}
}else if(for1==nums1Size){ // 情况2
count++;
num = nums2[for2++];
}else if(for2==nums2Size){ // 情况3
count++;
num = nums1[for1++];
}
}
an1 = num;
if(index1 == index2)
return an1; // 奇数总数
if((for1<nums1Size) && (for2<nums2Size)){ // 情况1
if(nums1[for1]<nums2[for2]){
count++;
an2 = nums1[for1++];
}else{
count++;
an2 = nums2[for2++];
}
}else if(for1==nums1Size){
count++;
an2 = nums2[for2++];
}else if(for2==nums2Size){
count++;
an2 = nums1[for1++];
}
return (an1+an2)/2;
}
238
// https://leetcode-cn.com/problems/product-of-array-except-self/
// 第一种解法，比较好，因为题目不让用除法，而第二种写之前没看清题目，所以用了除法
int* productExceptSelf(int* nums, int numsSize, int* returnSize){
returnSize[0] = numsSize;
int* answer = (int *)malloc(sizeof(int) * numsSize);
answer[0] = 1;
for(int i = 1;i <numsSize;i++){
answer[i] = answer[i-1] * nums[i-1];
}
int temp1=nums[numsSize-1],temp2;
nums[numsSize-1] = 1;
for(int i = numsSize-2; i>=0; i--){
temp2 = nums[i];
nums[i] = nums[i+1] * temp1;
temp1 = temp2;
}
for(int i = 0;i <numsSize;i++){
answer[i] = answer[i]*nums[i];
}
return answer;
}
// https://leetcode-cn.com/problems/product-of-array-except-self/
// 第二种解法，有问题，因为题目不让用除法
int* productExceptSelf(int* nums, int numsSize, int* returnSize){
returnSize[0] = numsSize;
int all = 1;
for(int i = 0;i <numsSize;i++){
all*= nums[i];
}
int* answer = (int *)malloc(sizeof(int) * numsSize);
for(int i = 0;i <numsSize;i++){
if(nums[i] != 0 )
answer[i] = all/nums[i];
else{
answer[i] =1;
for(int j =0; j<numsSize;j++){
if(j ==i)
continue;
answer[i] *= nums[j];
}
}
}
return answer;
}
974
// https://leetcode-cn.com/problems/subarray-sums-divisible-by-k/
int subarraysDivByK(int* A, int ASize, int K){
int fuzhu[ASize];
fuzhu[0] = A[0]%K;
if(fuzhu[0]<0)
fuzhu[0] += K;
int answer = 0;
int i;
for(i = 1; i < ASize; i++){
fuzhu[i] = (fuzhu[i-1]+A[i])%K;
if(fuzhu[i]<0)
fuzhu[i] += K;
}
int yu[K];
for(i = 0; i<K; i++){
yu[i] = 0;
}
for(i = 0; i < ASize; i++){
yu[fuzhu[i]]++;
}
yu[0] +=1;
for(i = 0; i<K; i++){
if(yu[i] >1)
answer+= yu[i]*(yu[i]-1)/2;
}
return answer;
}
628
// https://leetcode-cn.com/problems/maximum-product-of-three-numbers/
int maximumProduct(int* nums, int numsSize){
int a= nums[0],b = nums[1], c = nums[2];
void sort(void){ //小中大0
if(a>b){
int temp =a;
a = b;
b = temp;
}
if(b>c){
int temp = b;
b = c;
c =temp;
}
if(a>b){
int temp =a;
a = b;
b = temp;
}
}
if(numsSize==3)
return a*b*c;
// 获取最大数以及它的下标
int max = nums[0], maxindex = 0;
for(int i =0; i <numsSize; i++){
if(nums[i]>max){
max = nums[i];
maxindex = i;
}
}
// 获取完毕，对max进行判断
if(max == 0)
return 0;
if(max < 0 ){ // 全负情况
sort();
for(int i = 3;i <numsSize;i++){
if(nums[i]>a){
a = nums[i];
sort();
}
}
return (a*b*c);
}
// 最后一种情况，max>0
int z1=0,z2=0,f1=0,f2=0;
void zsort(void){ // 0小大
if(z1>z2){
int temp = z1;
z1 = z2;
z2 = temp;
}
}
void fsort(void){ // 小大0
if(f1>f2){
int temp = f1;
f1 = f2;
f2 = temp;
}
}
for(int i =0; i <numsSize; i++){
if(i == maxindex)
continue;
if(nums[i]>z1){
z1 = nums[i];
zsort();
}
if(nums[i]<f2){
f2 = nums[i];
fsort();
}
}
z1 = z1*z2;
f1 = f1*f2;
if(z1>f1)
return z1*max;
else
return f1*max;
}
206，剑指24
// 剑指24
// https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/
// 同主站206
// https://leetcode-cn.com/problems/reverse-linked-list/
struct ListNode* reverseList(struct ListNode* head){
if(head ==NULL)
return head;
struct ListNode * newhead = head, *p;
head = head->next;
newhead->next = NULL;
while(head){
p = head;
head = head->next;
p->next = newhead;
newhead = p;
}
return newhead;
}
剑指03
// 剑指03
// https://leetcode-cn.com/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/
int findRepeatNumber(int* nums, int numsSize){
int a;
int findone(int m){
int n = nums[m];
nums[m] = -1;
while((n!=-1)&&(nums[n] != n)){
m = nums[n];
nums[n] = n;
n = m;
}
return n;
}
for(int i = 0; i< numsSize; i++){
if(nums[i] != i){
a = i;
a = findone(a);
if(a!=-1)
return a;
}
}
return -1;
}
