Leetcode刷题---daily一练---编程基础从0到1题库

T1768：. - 力扣（LeetCode）

刷题时间：20240324

方法一：双指针法：

C语言题解：

char * mergeAlternately(char * word1, char * word2)
{
    char* mChar = (char*) malloc(strlen(word1) + strlen(word2) + 1);
    int i = 0, j = 0, k =0;
    for(;i < strlen(word1) && j < strlen(word2);)
    {
        mChar[k++] = word1[i++];
        mChar[k++] = word2[j++];
    }
    if (strlen(word1) > strlen(word2))
    {
        for(;i < strlen(word1); i++)
        {
            mChar[k++] = word1[i];
        }
    }
    else if(strlen(word1) < strlen(word2))
    {
        for(;j < strlen(word2); j++)
        {
            mChar[k++] = word2[j];
        }
    }
    mChar[k] = '\0';
    return mChar;
}

C语言题解方法二：

char * mergeAlternately(char * word1, char * word2){
    char *ans = (char *)calloc(1, strlen(word1) + strlen(word2) + 1), *p = ans;
    while (*word1 && *word2)
    {
        *p ++ = *word1 ++;
        *p ++ = *word2 ++;
    }
    while (*word1) *p ++ = *word1 ++;
    while (*word2) *p ++ = *word2 ++;
    return ans;
}

C++题解：

class Solution {
public:
    string mergeAlternately(string word1, string word2) {
        string retString;
        retString.reserve(word1.length() + word2.size());
        int i = 0, j = 0;
        while(i < word1.length() || j < word2.size()) {
            if (i < word1.size()) {
                retString.push_back(word1[i++]);
            }
            if (j < word2.length()) {
                retString.push_back(word2[j++]);
            }
        }
        return retString;
    }
};

java题解：

class Solution {
    public String mergeAlternately(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();

        int i = 0, j = 0;
        StringBuilder ans = new StringBuilder();
        while(i < m || j < n) {
            if (i < m) {
                ans.append(word1.charAt(i++));
            }
            if (j < n) {
                ans.append(word2.charAt(j++));
            }
        }
        return ans.toString();
    }
}

方法二：单指针法：

class Solu