力扣面试题（一）
1、给你两个字符串
word1和word2。请你从word1开始，通过交替添加字母来合并字符串。如果一个字符串比另一个字符串长，就将多出来的字母追加到合并后字符串的末尾。
char * mergeAlternately(char * word1, char * word2){int len1 = strlen(word1);int len2 = strlen(word2);char *merge_word = malloc(len1+len2+1);int i , j = 0;if(len1 <= len2){for(i = 0; i < len1; i++){merge_word[j++] = word1[i];merge_word[j++] = word2[i];}if(len1 < len2){for(i = len1; i < len2; i++)merge_word[j++] = word2[i];}}else{for(i = 0; i < len2; i++){merge_word[j++] = word1[i];merge_word[j++] = word2[i];}for(i = len2; i < len1; i++){merge_word[j++] = word1[i];}}merge_word[len1+len2] = '\0';return merge_word;
}给你一个数组
candies和一个整数extraCandies，其中candies[i]代表第i个孩子拥有的糖果数目。对每一个孩子，检查是否存在一种方案，将额外的
extraCandies个糖果分配给孩子们之后，此孩子有 最多 的糖果。注意，允许有多个孩子同时拥有 最多 的糖果数目。
bool* kidsWithCandies(int* candies, int candiesSize, int extraCandies, int* returnSize) {int max_candies = 0;int i = 0;for(i = 0; i < candiesSize; i++){if(max_candies < candies[i]){max_candies = candies[i];}}bool *candies_res = malloc(sizeof(bool)*candiesSize);*returnSize = candiesSize;for(i = 0 ; i < candiesSize; i++){if(candies[i] + extraCandies >= max_candies){candies_res[i] = true;}else{candies_res[i] = false;}}return candies_res;
}假设有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花不能种植在相邻的地块上，它们会争夺水源，两者都会死去。
给你一个整数数组
flowerbed表示花坛，由若干0和1组成，其中0表示没种植花，1表示种植了花。另有一个数n，能否在不打破种植规则的情况下种入n朵花？能则返回true，不能则返回false。示例 1：
输入：flowerbed = [1,0,0,0,1], n = 1 输出：true
bool canPlaceFlowers(int* flowerbed, int flowerbedSize, int n) {int unflower_count = 0;int i , j = 0;if(flowerbedSize ==1){if(!flowerbed[0]){j++;}}else{for(i = 0; i < flowerbedSize; i++){if(i == 0 && !flowerbed[i] && !flowerbed[i+1] ){j++;continue;}if(!flowerbed[i]){unflower_count ++;}else{unflower_count = 0;}if(unflower_count == 3){j++;i--;unflower_count = 0;}if(i == flowerbedSize-2 && flowerbed[i] == flowerbed[i+1] ){j++;break;}}}if(j >= n){return true;}return false;
}给你一个字符串
s，仅反转字符串中的所有元音字母，并返回结果字符串。元音字母包括
'a'、'e'、'i'、'o'、'u'，且可能以大小写两种形式出现不止一次。示例 1：
输入：s = "hello" 输出："holle"
char* reverseVowels(char* s) {int len = strlen(s);char * word = malloc(len*sizeof(char)+1);unsigned i , j, count= 0;//memset(word_location, -1, len*sizeof(int));unsigned * word_location = malloc(len*sizeof(unsigned));if(NULL == word_location){free(word);return NULL;}strcpy(word, s);for(i = 0 ; i < len; i ++){char lower_c = tolower(s[i]);if (lower_c == 'a' || lower_c == 'e' || lower_c == 'i' || lower_c == 'o' || lower_c == 'u') {word_location[count++] = i;} } if(count == 1){word[word_location[0]] = s[word_location[0]];}else{for(i = 0 ; i < count/2; i++){j = count -1-i;word[word_location[i]] = s[word_location[j]];word[word_location[j]] = s[word_location[i]];}}free(word_location);return word;
}