169. 多数元素 思想太妙了
中等题
228. 汇总区间 说好的简单题呢？？？
题解
int removeElement(int* nums, int numsSize, int val){
int left = 0, right = numsSize;
while (left < right) {
if (nums[left] == val) {//找到一个等于val的值 把它换到后面
nums[left] = nums[right - 1];
right--;//右端减一
} else {
left++;
}
}
return left;//left最终指向所有不等于val值的右边一个值
}
35. 搜索插入位置 基础的二分
int searchInsert(int* nums, int numsSize, int target){
int low = 0,high = numsSize - 1;
while(low <= high){
int mid = low + (high - low)/2;//根据中间值修改两个指针
if(nums[mid] > target) high = mid - 1;
if(nums[mid] == target) return mid;
if(nums[mid] < target) low = mid +1;
}
return low;
}
int maxSubArray(int* nums, int numsSize) {
int qian = 0, maxAns = nums[0];
for (int i = 0; i < numsSize; i++) {
qian = qian + nums[i]>nums[i]?qian + nums[i]:nums[i];//计算当前的最大前缀和
maxAns = maxAns>qian?maxAns:qian;
}
return maxAns;
}
int* plusOne(int* digits, int digitsSize, int* returnSize){
int count9 = 0;
for(int i = 0;i < digitsSize;i++) if(digits[i] == 9) count9++;
if(count9 == digitsSize) *returnSize = digitsSize + 1;//判断返回数组是否发生变化
else *returnSize = digitsSize;
int anssize = *returnSize,*ans = malloc(sizeof(int) * anssize);
int jinwei = 1;
for(int i = digitsSize-1;i >= 0;i--){
ans[--anssize] = (digits[i] + jinwei)%10;
jinwei = (digits[i] + jinwei)/10;
}
if(anssize) ans[--anssize] = 1;
return ans;
}
void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n){
int size = m + n;//从后往前合并 不会产生覆盖的问题
int min = -1000000001;
for(int i = m - 1,j = n - 1;i>=0||j>=0;){
if(i<0) nums1[--size] = nums2[j--];
else if(j<0||nums1[i] > nums2[j]) nums1[--size] = nums1[i--];
else nums1[--size] = nums2[j--];
}
}
int maxProfit(int* prices, int pricesSize){
int minprice = 1000000000, maxprofit = 0;//找到最低点买入 高点卖出
for (int i = 0;i<pricesSize;i++) {
if((prices[i] - minprice) > maxprofit) maxprofit = (prices[i] - minprice);
if(minprice > prices[i]) minprice = prices[i];
}
return maxprofit;
}
int* twoSum(int* numbers, int numbersSize, int target, int* returnSize){
int hash[2010];//创建hash表
memset(hash,-1,sizeof(hash));//初始为-1，记录位置
*returnSize = 2;
int *ans = malloc(sizeof(int) * 2);
for(int i = 0;i < numbersSize;i++)
if(hash[target - numbers[i]+1000] != -1){
ans[0] = hash[target - numbers[i]+1000];
ans[1] = i + 1;
return ans;
}
else
hash[numbers[i] + 1000] = i + 1;//+1000保证映射到数组内
return 0;
}
169. 多数元素 由于超过一半 每一那一个最多次数的元素和一个非最多元素抵消 最后剩下的只能是最多元素
int majorityElement(int* nums, int numsSize){
int maxnum = 0,maxtime = 0;//最多元素 以及最多元素当前的次数
for(int i = 0;i < numsSize;i++){
if(maxtime == 0){
maxnum = nums[i];//没有最多元素 选定当前元素为最多元素
maxtime++;
}
else if(nums[i] == maxnum) maxtime++;//还是最多元素 增加计数
else maxtime--;//不是最多计数 抵消一个?
}
return maxnum;
}
int cmp(int *a,int *b){
return *a > *b;
}
bool containsDuplicate(int* nums, int numsSize){
qsort(nums,numsSize,sizeof(int),cmp);//排个序直接看被。。
for(int i = 1;i < numsSize;i++)
if(nums[i] == nums[i-1]) return true;
return false;
}
void moveZeroes(int* nums, int numsSize){
int i = 0;
while(i<numsSize&&nums[i++] != 0);
if(i == numsSize) return ;
i--;
for(int j = i + 1;j < numsSize;j++){
if(nums[j] != 0) nums[i++] = nums[j];
}
for(;i<numsSize;i++) nums[i] = 0;
}
typedef struct {
int sum[20000];
int size;
} NumArray;
NumArray* numArrayCreate(int* nums, int numsSize) {
NumArray * ans= malloc(sizeof(NumArray));
ans->size = numsSize;
ans->sum[0] = 0;
for(int i = 0;i < numsSize;i++)
ans->sum[i+1] = ans->sum[i] + nums[i];
return ans;
}
int numArraySumRange(NumArray* obj, int left, int right) {
return obj->sum[right+1] - obj->sum[left];
}
void numArrayFree(NumArray* obj) {
free(obj);
}
void rever(char *s){
int len = strlen(s);
for(int i = 0;i < len/2;i++){
s[i] = s[i] ^ s[len - 1 - i];
s[len - 1 - i] = s[i] ^ s[len - 1 - i];
s[i] = s[i] ^ s[len - 1 - i];
}
}
char *numtostr(int n){
char * ans = malloc(sizeof(char)*12);
if(n==0){
ans[0] = '0',ans[1] = 0;
return ans;
}
int anssize = 0,flag = 1;
if(n > 0)
flag = 0,n = -n;//转化为负数
while(n){
ans[anssize++] = '0' - n % 10;//处理数字
n/=10;
}
if(flag) ans[anssize++] = '-';
ans[anssize++] = 0;
rever(ans);
return ans;
}
char ** summaryRanges(int* nums, int numsSize, int* returnSize){
char** ans = malloc(sizeof(char*)*numsSize);
char fuhao[3] = {"->"};
int anssize = 0;
for(int i = 0;i<numsSize;i++){
int temp = i;i++;
while(i<numsSize && (long long )nums[i]-(long long )nums[temp] == i - temp) i++;
i--;
if(i == temp)
ans[anssize++] = numtostr(nums[temp]);
else{
char * anstemp = malloc(sizeof(char)*25);
anstemp[0] = 0;
strcat(anstemp,numtostr(nums[temp]));
strcat(anstemp,fuhao);
strcat(anstemp,numtostr(nums[i]));
ans[anssize++] = anstemp;
}
}
*returnSize = anssize;
return ans;
}
char ** findWords(char ** words, int wordsSize, int* returnSize){
char **ans = malloc(sizeof(char*)*wordsSize);
int anssize = 0;
char s[3][20]={"qwertyuiop","asdfghjkl","zxcvbnm"};
bool hash[3][26];
for(int i = 0;i<3;i++){
memset(hash[i],0,sizeof(hash[i]));
for(int j =0;j < strlen(s[i]);j++)
hash[i][s[i][j] - 'a'] = true;
}
for(int i = 0;i<wordsSize;i++){
if(!words[i][0]) continue;
int j =-1,k;
while(++j >-1){ //寻找第几排
if(words[i][0]<='Z'){
if(hash[j][words[i][0] - 'A']) break;
}
else
if(hash[j][words[i][0] - 'a']) break;
}
for(k = 1;words[i][k];k++)
if(words[i][k] <='Z'){
if(!hash[j][words[i][k] - 'A']) break;
}
else
if(!hash[j][words[i][k] - 'a']) break;
printf("%d %d ",j,k);
if(!words[i][k]) ans[anssize++] = words[i];
}
*returnSize = anssize;
return ans;
}