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feat: Revised the book (#978)
* Sync recent changes to the revised Word. * Revised the preface chapter * Revised the introduction chapter * Revised the computation complexity chapter * Revised the chapter data structure * Revised the chapter array and linked list * Revised the chapter stack and queue * Revised the chapter hashing * Revised the chapter tree * Revised the chapter heap * Revised the chapter graph * Revised the chapter searching * Reivised the sorting chapter * Revised the divide and conquer chapter * Revised the chapter backtacking * Revised the DP chapter * Revised the greedy chapter * Revised the appendix chapter * Revised the preface chapter doubly * Revised the figures
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@ -13,7 +13,7 @@ void bucketSort(vector<float> &nums) {
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vector<vector<float>> buckets(k);
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// 1. 将数组元素分配到各个桶中
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for (float num : nums) {
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// 输入数据范围 [0, 1),使用 num * k 映射到索引范围 [0, k-1]
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// 输入数据范围为 [0, 1),使用 num * k 映射到索引范围 [0, k-1]
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int i = num * k;
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// 将 num 添加进桶 bucket_idx
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buckets[i].push_back(num);
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@ -36,7 +36,7 @@ void heapSort(vector<int> &nums) {
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}
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// 从堆中提取最大元素,循环 n-1 轮
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for (int i = nums.size() - 1; i > 0; --i) {
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// 交换根节点与最右叶节点(即交换首元素与尾元素)
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// 交换根节点与最右叶节点(交换首元素与尾元素)
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swap(nums[0], nums[i]);
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// 以根节点为起点,从顶至底进行堆化
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siftDown(nums, i, 0);
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@ -18,7 +18,7 @@ class QuickSort {
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/* 哨兵划分 */
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static int partition(vector<int> &nums, int left, int right) {
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// 以 nums[left] 作为基准数
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// 以 nums[left] 为基准数
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int i = left, j = right;
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while (i < j) {
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while (i < j && nums[j] >= nums[left])
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@ -73,7 +73,7 @@ class QuickSortMedian {
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int med = medianThree(nums, left, (left + right) / 2, right);
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// 将中位数交换至数组最左端
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swap(nums, left, med);
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// 以 nums[left] 作为基准数
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// 以 nums[left] 为基准数
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int i = left, j = right;
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while (i < j) {
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while (i < j && nums[j] >= nums[left])
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@ -112,7 +112,7 @@ class QuickSortTailCall {
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/* 哨兵划分 */
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static int partition(vector<int> &nums, int left, int right) {
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// 以 nums[left] 作为基准数
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// 以 nums[left] 为基准数
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int i = left, j = right;
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while (i < j) {
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while (i < j && nums[j] >= nums[left])
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@ -132,7 +132,7 @@ class QuickSortTailCall {
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while (left < right) {
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// 哨兵划分操作
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int pivot = partition(nums, left, right);
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// 对两个子数组中较短的那个执行快排
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// 对两个子数组中较短的那个执行快速排序
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if (pivot - left < right - pivot) {
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quickSort(nums, left, pivot - 1); // 递归排序左子数组
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left = pivot + 1; // 剩余未排序区间为 [pivot + 1, right]
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@ -14,7 +14,7 @@ int digit(int num, int exp) {
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/* 计数排序(根据 nums 第 k 位排序) */
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void countingSortDigit(vector<int> &nums, int exp) {
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// 十进制的位范围为 0~9 ,因此需要长度为 10 的桶
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// 十进制的位范围为 0~9 ,因此需要长度为 10 的桶数组
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vector<int> counter(10, 0);
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int n = nums.size();
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// 统计 0~9 各数字的出现次数
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