【C语言入门】第四天

【例题1】2325. 解密消息 - 力扣（LeetCode）

int isTrue(char c,char chs[26],int pos){
    int i;
    for(i=0;i<pos;i++){
        if(c == chs[i]){
            return 0;
        }
    }
    return 1;
}

char lookForChar(char c, char chs[26]){
    for(int i = 0;i<26;i++){
        if(c == chs[i]){
            return (char)('a'+i);
        }
    }
    return ' ';
}

char * decodeMessage(char * key, char * message){
    char* chs = (char*)malloc(sizeof(char)*27);
    int cur = 0;int pos = 0;
    while(*(key+cur) != '\0' ){
        char ch = *(key+cur);
        if( ch<='z' && ch >= 'a' && isTrue(ch,chs,pos)){
            *(chs+pos) = ch;
            pos++;
        }
        cur++;
    }
    char* newStr = (char*)malloc(sizeof(char)*2001);
    cur = 0;
    while(*(message+cur) != '\0'){
        if (*(message+cur) != ' ') {    
            *(newStr+cur) = lookForChar(*(message+cur),chs);
        }else{
            *(newStr+cur) = ' ';
        }
        cur++;
    }
    *(newStr+cur) = '\0';
    return newStr;
    
}

【例题2】1379. 找出克隆二叉树中的相同节点 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
        if(target->val == cloned->val){
            return cloned;
        }
        if(cloned->left != NULL){
            original = getTargetCopy(original,cloned->left,target);
        }
        if(cloned->right != NULL){
            original = getTargetCopy(original,cloned->right,target);
        }
        return original;
    }
};

【例题3】2469. 温度转换 - 力扣（LeetCode）

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
double* convertTemperature(double celsius, int* returnSize){
    double* ans = (double*)malloc(sizeof(double)*2);
    *(ans+0) = celsius + 273.15;
    *(ans+1) = celsius*1.80 + 32.00;
    *returnSize = 2;
     return ans;
}

【例题4】LCR 189. 设计机械累加器 - 力扣（LeetCode）

int mechanicalAccumulator(int target) {
    return target == 1 ? 1 : mechanicalAccumulator(target-1)+target;
}

【例题5】371. 两整数之和 - 力扣（LeetCode）

int getSum(int a, int b){
    while(b!=0){
        unsigned int temp = (unsigned int) (a&b)<<1; // 得到进位
        a = a^b; // 获得无进位加法结果
        b = temp;
    }
    return a;
}