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Re-implement merge sort function. (#888)

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Yudong Jin
2023-10-26 02:56:33 +08:00
committed by GitHub
parent 7822bf9cd4
commit e441ee4e35
8 changed files with 186 additions and 181 deletions
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46
codes/javascript/chapter_sorting/merge_sort.js
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@ -5,33 +5,33 @@
*/
/* 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]
function merge(nums, left, mid, right) {
// 初始化辅助数组
let tmp = nums.slice(left, right + 1);
// 左子数组的起始索引和结束索引
let leftStart = left - left,
leftEnd = mid - left;
// 右子数组的起始索引和结束索引
let rightStart = mid + 1 - left,
rightEnd = right - left;
// i, j 分别指向左子数组、右子数组的首元素
let i = leftStart,
j = rightStart;
// 通过覆盖原数组 nums 来合并左子数组和右子数组
for (let k = left; k <= right; k++) {
if (i > leftEnd) {
// 若“左子数组已全部合并完”，则选取右子数组元素，并且 j++
nums[k] = tmp[j++];
} else if (j > rightEnd || tmp[i] <= 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++
nums[k] = tmp[i++];
// 左子数组区间 [left, mid], 右子数组区间 [mid+1, right]
// 创建一个临时数组 tmp ，用于存放合并后的结果
const tmp = new Array(right - left + 1);
// 初始化左子数组和右子数组的起始索引
let i = left,
j = mid + 1,
k = 0;
// 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
while (i <= mid && j <= right) {
if (nums[i] <= nums[j]) {
tmp[k++] = nums[i++];
} else {
// 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++
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];
}
}
/* 归并排序 */