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添加 0077.组合.md C语言版本
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@ -435,6 +435,59 @@ func backtrack(n,k,start int,track []int){
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}
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```
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C:
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```c
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int* path;
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int pathTop;
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int** ans;
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int ansTop;
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void backtracking(int n, int k,int startIndex) {
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//当path中元素个数为k个时,我们需要将path数组放入ans二维数组中
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if(pathTop == k) {
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//path数组为我们动态申请,若直接将其地址放入二维数组,path数组中的值会随着我们回溯而逐渐变化
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//因此创建新的数组存储path中的值
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int* temp = (int*)malloc(sizeof(int) * k);
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int i;
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for(i = 0; i < k; i++) {
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temp[i] = path[i];
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}
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ans[ansTop++] = temp;
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return ;
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}
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int j;
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for(j = startIndex; j <=n ;j++) {
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//将当前结点放入path数组
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path[pathTop++] = j;
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//进行递归
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backtracking(n, k, j + 1);
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//进行回溯,将数组最上层结点弹出
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pathTop--;
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}
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}
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int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
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//path数组存储符合条件的结果
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path = (int*)malloc(sizeof(int) * k);
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//ans二维数组存储符合条件的结果数组的集合。(数组足够大,避免极端情况)
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ans = (int**)malloc(sizeof(int*) * 10000);
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pathTop = ansTop = 0;
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//回溯算法
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backtracking(n, k, 1);
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//最后的返回大小为ans数组大小
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*returnSize = ansTop;
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//returnColumnSizes数组存储ans二维数组对应下标中一维数组的长度(都为k)
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*returnColumnSizes = (int*)malloc(sizeof(int) *(*returnSize));
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int i;
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for(i = 0; i < *returnSize; i++) {
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(*returnColumnSizes)[i] = k;
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}
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//返回ans二维数组
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return ans;
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}
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```
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-----------------------
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* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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