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Merge pull request #730 from KingArthur0205/remote
添加 0059.螺旋矩阵II.md C语言版本 以及 添加 0216.组合总和II.md C语言版本
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@ -469,6 +469,65 @@ class Solution {
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}
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```
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C:
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```c
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int** generateMatrix(int n, int* returnSize, int** returnColumnSizes){
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//初始化返回的结果数组的大小
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*returnSize = n;
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*returnColumnSizes = (int*)malloc(sizeof(int) * n);
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//初始化返回结果数组ans
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int** ans = (int**)malloc(sizeof(int*) * n);
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int i;
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for(i = 0; i < n; i++) {
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ans[i] = (int*)malloc(sizeof(int) * n);
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(*returnColumnSizes)[i] = n;
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}
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//设置每次循环的起始位置
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int startX = 0;
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int startY = 0;
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//设置二维数组的中间值,若n为奇数。需要最后在中间填入数字
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int mid = n / 2;
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//循环圈数
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int loop = n / 2;
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//偏移数
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int offset = 1;
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//当前要添加的元素
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int count = 1;
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while(loop) {
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int i = startX;
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int j = startY;
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//模拟上侧从左到右
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for(; j < startY + n - offset; j++) {
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ans[startX][j] = count++;
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}
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//模拟右侧从上到下
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for(; i < startX + n - offset; i++) {
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ans[i][j] = count++;
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}
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//模拟下侧从右到左
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for(; j > startY; j--) {
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ans[i][j] = count++;
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}
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//模拟左侧从下到上
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for(; i > startX; i--) {
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ans[i][j] = count++;
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}
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//偏移值每次加2
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offset+=2;
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//遍历起始位置每次+1
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startX++;
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startY++;
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loop--;
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}
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//若n为奇数需要单独给矩阵中间赋值
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if(n%2)
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ans[mid][mid] = count;
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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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@ -392,7 +392,63 @@ var combinationSum3 = function(k, n) {
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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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int getPathSum() {
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int i;
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int sum = 0;
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for(i = 0; i < pathTop; i++) {
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sum += path[i];
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}
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return sum;
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}
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void backtracking(int targetSum, int k, int sum, int startIndex) {
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if(pathTop == k) {
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if(sum == targetSum) {
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int* tempPath = (int*)malloc(sizeof(int) * k);
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int j;
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for(j = 0; j < k; j++)
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tempPath[j] = path[j];
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ans[ansTop++] = tempPath;
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}
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// 如果path.size() == k 但sum != targetSum 直接返回
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return;
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}
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int i;
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//从startIndex开始遍历,一直遍历到9
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for (i = startIndex; i <= 9; i++) {
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sum += i; // 处理
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path[pathTop++] = i; // 处理
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backtracking(targetSum, k, sum, i + 1); // 注意i+1调整startIndex
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sum -= i; // 回溯
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pathTop--;; // 回溯
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}
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}
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int** combinationSum3(int k, int n, int* returnSize, int** returnColumnSizes){
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//初始化辅助变量
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path = (int*)malloc(sizeof(int) * k);
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ans = (int**)malloc(sizeof(int*) * 20);
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pathTop = ansTop = 0;
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backtracking(n, k, 0, 1);
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//设置返回的二维数组中元素个数为ansTop
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*returnSize = ansTop;
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//设置二维数组中每个元素个数的大小为k
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*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
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int i;
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for(i = 0; i < ansTop; i++) {
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(*returnColumnSizes)[i] = k;
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}
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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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