LeetCode-4Sum(九月千题鬼之历练 1/1000, 指针数组的初始化和调用)

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d =
target? Find all unique quadruplets in the array which gives the sum of target.

Note:

Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
The solution set must not contain duplicate quadruplets.

For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

A solution set is:
(-1,  0, 0, 1)
(-2, -1, 1, 2)
(-2,  0, 0, 2)

给一个数组，nums, 有numsSize个数，问其中有多少种4个数字的subset之和等于target。

/**
* Return an array of arrays of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int** fourSum(int* nums, int numsSize, int target, int* returnSize)

题目的tricks：

1) 数组元素个数可能小于4

2) 相同数字组成的set需要去重，例如: nums = [0, 0, 0, 0, 0, 0, 0]，target=0, 仅有一组解，[0, 0, 0, 0]。

3) 由于结果的个数不可确定，指针数组的初始化的大小不确定，在最后需要mencpy。所以在最开始，初始化的时候，malloc了10000个sizeof(int*).

解题思路：

1）首先，将nums从小到大排序

2）取四个数，a, b, c, d 遍历一组4个数的下标，用暴力循环确定a -> 0 ~ numsSize - 4, b -> a ~ numsSize

for (a = 0; a <= numsSize - 4; ){
for (b = a + 1; b <= numsSize - 3; ){
..............
while (nums == nums[b + 1] && b <= numsSize - 3){
b++;
}
b++;
}
while (nums[a] == nums[a + 1] && a <= numsSize - 4){
a++;
}
a++;
}

3) c = b + 1, d = numsSize - 1

也就是说，保持始终a < b < c < d.

4）当nums[a] + nums[b] + nums[c] + nums[d] > target 的时候，说明nums[d]太大，d--;

当nums[a] + nums[b] + nums[c] + nums[d] < target的时候，说明nums[c]太小, c++;

当nums[a]
+ nums[b] + nums[c] + nums[d] = target 的时候，(nums[a], nums[b], nums[c], nums[d]）为一组解，保存；

Tips: 当改变a, b, c, d的时候，确保改变之后对应的数值和之前的值不相同，以免出现重复的解。

[b]指针数组的初始化和调用：

变量：int **point

初始化：point = (int**)malloc(sizeof(int*) * NUM)

赋值：int *block = (int*)malloc(sizeof(int*) * BLOCK_NUM);

block[0] = 3; block[1] = 4; block[2] = 5;

point[i] = block;

调用：point[0][0] = 3;

代码如下：

#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <cstring>

#define MAX_NUM 10000

/**
* Return an array of arrays of size *returnSize.
* Note: The returned array must be malloced, assume caller calls free().
*/
int** fourSum(int* nums, int numsSize, int target, int* returnSize) {
if (numsSize < 4){
return NULL;
}

int a, b, c, d;
int **result;
int **tmp_result;
int *tmp_loc;
int i, j;
int sum, tmp;

*returnSize = 0;
result = (int **)malloc(sizeof(int *) * (MAX_NUM));

for (i = 0; i < numsSize; i++){
for (j = 0; j < numsSize - i - 1; j++){
if (nums[j] > nums[j + 1]){
tmp = nums[j];
nums[j] = nums[j + 1];
nums[j + 1] = tmp;
}
}
}

for (a = 0; a <= numsSize - 4; ){
for (b = a + 1; b <= numsSize - 3; ){
for (c = b + 1, d = numsSize - 1; c < d; ){
sum = nums[a] + nums[b] + nums[c] + nums[d];
if (sum < target){
while (nums[c] == nums[c + 1] && c < d){
c++;
}
c++;
}
else if (sum > target){
while (nums[d] == nums[d - 1] && c < d){
d--;
}
d--;
}
else if (sum == target){
tmp_loc = (int*) malloc(sizeof(int) * 4);
tmp_loc[0] = nums[a];
tmp_loc[1] = nums[b];
tmp_loc[2] = nums[c];
tmp_loc[3] = nums[d];

result[*returnSize] = tmp_loc;

*returnSize = *returnSize + 1;

while (nums[c] == nums[c + 1] && c < d){
c++;
}
c++;
while (nums[d] == nums[d - 1] && c < d){
d--;
}
d--;
}

}
while (nums[b] == nums[b + 1] && b <= numsSize - 3){
b++;
}
b++;
}
while (nums[a] == nums[a + 1] && a <= numsSize - 4){
a++;
}
a++;
}

tmp_result = (int **)malloc(sizeof(int *) * (*returnSize));
memcpy(tmp_result, result, *returnSize * sizeof(int*));

free(result);

return tmp_result;
}

int main(void){
int arr[] = {1,-2,-5,-4,-3,3,3,5};
int target = -11;
int size = 8;
int i, j;
int returnSize;
int **result;

result = fourSum(arr, size, target, &returnSize);

printf("%d\n", returnSize);

for (i = 0; i < returnSize; i++){
for (j = 0; j < 4; j++){
printf("%d, ", result[i][j]);
}
printf("\n");
}
return 0;
}