代码随想录算法训练营day27 39. 组合总和 40.组合总和II 131.分割回文串

组合总和

不是很懂数组和指针的部分，回溯的部分逻辑搞懂了，其间那个另写temp数组防止path变化的步骤也涉及到了数组和指针，这部分要抽出时间来学。
和前两天的一样，不过递归部分自己也递归进去而不是从下一个开始。还有就是每组答案的长度不一样要一个length来记录，又是数组上的知识。

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int* path;
int pathTop;
int** ans;
int ansTop;
//记录每一个和等于target的path数组长度
int* length;

void backTracking(int target, int index, int* candidates, int candidatesSize, int sum) {
    if(sum>=target)
    {
        if(sum == target)
        {
        int* temp = (int*)malloc(sizeof(int)*pathTop);
        int j;
        for(j=0;j<pathTop;j++)
        {
            temp[j] = path[j];
        }
        ans[ansTop] = temp;
        length[ansTop++] = pathTop;
        
        }
        return ;
    }
    int i;
    for(i = index;i < candidatesSize;i++)
    {
        sum+=candidates[i];
        path[pathTop++] = candidates[i];
        backTracking(target,i,candidates,candidatesSize,sum);
        sum-=candidates[i];
        pathTop--;
    }
}
int** combinationSum(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes){
    //初始化变量
    path = (int*)malloc(sizeof(int) * 50);
    ans = (int**)malloc(sizeof(int*) * 200);
    length = (int*)malloc(sizeof(int) * 200);
    ansTop = pathTop = 0;
    backTracking(target, 0, candidates, candidatesSize, 0);

    //设置返回的数组大小
    *returnSize = ansTop;
    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
    int i;
    for(i = 0; i < ansTop; i++) {
        (*returnColumnSizes)[i] = length[i];
    }
    return ans;
}

组合总和II

查重，前后元素一样后元素能满足的组合条件前元素一定能满足

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int* path;
int pathTop;
int** ans;
int ansTop;
//记录ans中每一个一维数组的大小
int* length;
int cmp(const void* a1, const void* a2) {
    return *((int*)a1) - *((int*)a2);
}

void backTracking(int* candidates, int candidatesSize,  int target, int sum, int startIndex){
    if(sum >= target) {
        //若sum等于target，复制当前path进入
        if(sum == target) {
            int* tempPath = (int*)malloc(sizeof(int) * pathTop);
            int j;
            for(j = 0; j < pathTop; j++) {
                tempPath[j] = path[j];
            }
            length[ansTop] = pathTop;
            ans[ansTop++] = tempPath;
        }
        return ;
    }
    int i;
    for(i = startIndex;i<candidatesSize;i++)
    {
        if(i>startIndex&&candidates[i]==candidates[i-1])
        {
            continue;//再往下就重了，跳过
        }
        path[pathTop++] = candidates[i];
        sum += candidates[i];
        backTracking(candidates, candidatesSize, target, sum, i + 1);
        //回溯
        sum -= candidates[i];
        pathTop--;

    }
}

int** combinationSum2(int* candidates, int candidatesSize, int target, int* returnSize, int** returnColumnSizes){
    path = (int*)malloc(sizeof(int) * 50);
    ans = (int**)malloc(sizeof(int*) * 100);
    length = (int*)malloc(sizeof(int) * 100);
    pathTop = ansTop = 0;
    //快速排序candidates，让相同元素挨到一起
    qsort(candidates, candidatesSize, sizeof(int), cmp);

    backTracking(candidates, candidatesSize, target, 0, 0);

    *returnSize = ansTop;
    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
    int i;
    for(i = 0; i < ansTop; i++) {
        (*returnColumnSizes)[i] = length[i];
    }
    return ans;
}

分割回文串(写不出来）

5.18:逻辑方面搞懂了还是数组方面的问题实在搞不懂里扣的那些return指针

char** path;
int pathTop;
char*** ans;
int ansTop = 0;
int* ansSize;

//将path中的字符串全部复制到ans中
void copy() {
    //创建一个临时tempPath保存path中的字符串
    char** tempPath = (char**)malloc(sizeof(char*) * pathTop);
    int i;
    for(i = 0; i < pathTop; i++) {
        tempPath[i] = path[i];
    }
    //保存tempPath
    ans[ansTop] = tempPath;
    //将当前path的长度（pathTop）保存在ansSize中
    ansSize[ansTop++] = pathTop;
}

//判断字符串是否为回文字符串
bool isPalindrome(char* str, int startIndex, int endIndex) {
    //双指针法：当endIndex（右指针）的值比startIndex（左指针）大时进行遍历
    while(endIndex >= startIndex) {
        //若左指针和右指针指向元素不一样，返回False
        if(str[endIndex--] != str[startIndex++])
            return 0;
    }
    return 1;
}

//切割从startIndex到endIndex子字符串
char* cutString(char* str, int startIndex, int endIndex) {
    //开辟字符串的空间
    char* tempString = (char*)malloc(sizeof(char) * (endIndex - startIndex + 2));
    int i;
    int index = 0;
    //复制子字符串
    for(i = startIndex; i <= endIndex; i++)
        tempString[index++] = str[i];
    //用'\0'作为字符串结尾
    tempString[index] = '\0';
    return tempString;
}

void backTracking(char* str, int strLen,  int startIndex) {
    if(startIndex >= strLen) {
        //将path拷贝到ans中
        copy();
        return ;
    }

    int i;
    for(i = startIndex; i < strLen; i++) {
        //若从subString到i的子串是回文字符串，将其放入path中
        if(isPalindrome(str, startIndex, i)) {
            path[pathTop++] = cutString(str, startIndex, i);
        }
        //若从startIndex到i的子串不为回文字符串，跳过这一层 
        else {
            continue;
        }
        //递归判断下一层
        backTracking(str, strLen, i + 1);
        //回溯，将path中最后一位元素弹出
        pathTop--;
    }
}

char*** partition(char* s, int* returnSize, int** returnColumnSizes){
    int strLen = strlen(s);
    //因为path中的字符串最多为strLen个（即单个字符的回文字符串），所以开辟strLen个char*空间
    path = (char**)malloc(sizeof(char*) * strLen);
    //存放path中的数组结果
    ans = (char***)malloc(sizeof(char**) * 40000);
    //存放ans数组中每一个char**数组的长度
    ansSize = (int*)malloc(sizeof(int) * 40000);
    ansTop = pathTop = 0;

    //回溯函数
    backTracking(s, strLen, 0);

    //将ansTop设置为ans数组的长度
    *returnSize = ansTop;
    //设置ans数组中每一个数组的长度
    *returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
    int i;
    for(i = 0; i < ansTop; ++i) {
        (*returnColumnSizes)[i] = ansSize[i];
    }
    return ans;
}