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refactor(dart): Re-implement merge sort function (#896)

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liuyuxin 2023-10-27 16:00:26 +08:00 committed by GitHub
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codes/dart/chapter_sorting/merge_sort.dart
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@ -5,28 +5,29 @@
*/ */
/* 合并左子数组和右子数组 */ /* 合并左子数组和右子数组 */
// 左子数组区间 [left, mid]
// 右子数组区间 [mid + 1, right]
void merge(List<int> nums, int left, int mid, int right) { void merge(List<int> nums, int left, int mid, int right) {
// 初始化辅助数组 // 左子数组区间 [left, mid], 右子数组区间 [mid+1, right]
List<int> tmp = nums.sublist(left, right + 1); // 创建一个临时数组 tmp ，用于存放合并后的结果
// 左子数组的起始索引和结束索引 List<int> tmp = List.filled(right - left + 1, 0);
int leftStart = left - left, leftEnd = mid - left; // 初始化左子数组和右子数组的起始索引
// 右子数组的起始索引和结束索引 int i = left, j = mid + 1, k = 0;
int rightStart = mid + 1 - left, rightEnd = right - left; // 当左右子数组都还有元素时，比较并将较小的元素复制到临时数组中
// i, j 分别指向左子数组、右子数组的首元素 while (i <= mid && j <= right) {
int i = leftStart, j = rightStart; if (nums[i] <= nums[j])
// 通过覆盖原数组 nums 来合并左子数组和右子数组 tmp[k++] = nums[i++];
for (int k = left; k <= right; k++) {
// 若“左子数组已全部合并完”，则选取右子数组元素，并且 j++
if (i > leftEnd)
nums[k] = tmp[j++];
// 否则，若“右子数组已全部合并完”或“左子数组元素 <= 右子数组元素”，则选取左子数组元素，并且 i++
else if (j > rightEnd || tmp[i] <= tmp[j])
nums[k] = tmp[i++];
// 否则，若“左右子数组都未全部合并完”且“左子数组元素 > 右子数组元素”，则选取右子数组元素，并且 j++
else else
nums[k] = tmp[j++]; tmp[k++] = nums[j++];
}
// 将左子数组和右子数组的剩余元素复制到临时数组中
while (i <= mid) {
tmp[k++] = nums[i++];
}
while (j <= right) {
tmp[k++] = nums[j++];
}
// 将临时数组 tmp 中的元素复制回原数组 nums 的对应区间
for (k = 0; k < tmp.length; k++) {
nums[left + k] = tmp[k];
} }
} }