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Commit 26650aa

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feat: add solutions to lc problem: No.0632 (#3672)
1 parent c37ef9b commit 26650aa

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‎solution/0600-0699/0632.Smallest Range Covering Elements from K Lists/README.md‎

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<pre>
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<strong>输入:</strong>nums = [[4,10,15,24,26], [0,9,12,20], [5,18,22,30]]
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<strong>输出:</strong>[20,24]
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<strong>解释:</strong>
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<strong>解释:</strong>
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列表 1:[4, 10, 15, 24, 26],24 在区间 [20,24] 中。
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列表 2:[0, 9, 12, 20],20 在区间 [20,24] 中。
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列表 3:[5, 18, 22, 30],22 在区间 [20,24] 中。
@@ -269,4 +269,80 @@ impl Solution {
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<!-- solution:end -->
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<!-- solution:start -->
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### 方法二:优先队列(小根堆)
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<!-- tabs:start -->
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#### TypeScript
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```ts
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const smallestRange = (nums: number[][]): number[] => {
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const pq = new MinPriorityQueue<[number, number, number]>({ priority: ([x]) => x });
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const n = nums.length;
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let [l, r, max] = [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
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for (let j = 0; j < n; j++) {
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const x = nums[j][0];
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pq.enqueue([x, j, 0]);
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max = Math.max(max, x);
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}
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while (pq.size() === n) {
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const [min, j, i] = pq.dequeue().element;
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if (max - min < r - l) {
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[l, r] = [min, max];
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}
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const iNext = i + 1;
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if (iNext < nums[j].length) {
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const next = nums[j][iNext];
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pq.enqueue([next, j, iNext]);
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max = Math.max(max, next);
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}
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}
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return [l, r];
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};
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```
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#### JavaScript
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```js
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const smallestRange = nums => {
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const pq = new MinPriorityQueue({ priority: ([x]) => x });
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const n = nums.length;
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let [l, r, max] = [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
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for (let j = 0; j < n; j++) {
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const x = nums[j][0];
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pq.enqueue([x, j, 0]);
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max = Math.max(max, x);
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}
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while (pq.size() === n) {
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const [min, j, i] = pq.dequeue().element;
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if (max - min < r - l) {
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[l, r] = [min, max];
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}
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const iNext = i + 1;
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if (iNext < nums[j].length) {
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const next = nums[j][iNext];
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pq.enqueue([next, j, iNext]);
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max = Math.max(max, next);
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}
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}
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return [l, r];
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/0600-0699/0632.Smallest Range Covering Elements from K Lists/README_EN.md‎

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@@ -257,4 +257,80 @@ impl Solution {
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<!-- solution:end -->
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<!-- solution:start -->
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### Solution 2: Priority Queue (Heap)
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<!-- tabs:start -->
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#### TypeScript
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```ts
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const smallestRange = (nums: number[][]): number[] => {
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const pq = new MinPriorityQueue<[number, number, number]>({ priority: ([x]) => x });
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const n = nums.length;
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let [l, r, max] = [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
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for (let j = 0; j < n; j++) {
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const x = nums[j][0];
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pq.enqueue([x, j, 0]);
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max = Math.max(max, x);
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}
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while (pq.size() === n) {
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const [min, j, i] = pq.dequeue().element;
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if (max - min < r - l) {
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[l, r] = [min, max];
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}
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const iNext = i + 1;
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if (iNext < nums[j].length) {
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const next = nums[j][iNext];
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pq.enqueue([next, j, iNext]);
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max = Math.max(max, next);
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}
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}
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return [l, r];
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};
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```
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#### JavaScript
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```js
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const smallestRange = nums => {
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const pq = new MinPriorityQueue({ priority: ([x]) => x });
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const n = nums.length;
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let [l, r, max] = [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
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for (let j = 0; j < n; j++) {
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const x = nums[j][0];
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pq.enqueue([x, j, 0]);
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max = Math.max(max, x);
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}
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while (pq.size() === n) {
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const [min, j, i] = pq.dequeue().element;
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if (max - min < r - l) {
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[l, r] = [min, max];
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}
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const iNext = i + 1;
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if (iNext < nums[j].length) {
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const next = nums[j][iNext];
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pq.enqueue([next, j, iNext]);
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max = Math.max(max, next);
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}
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}
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return [l, r];
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};
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->
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const smallestRange = nums => {
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const pq = new MinPriorityQueue({ priority: ([x]) => x });
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const n = nums.length;
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let [l, r, max] = [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
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for (let j = 0; j < n; j++) {
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const x = nums[j][0];
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pq.enqueue([x, j, 0]);
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max = Math.max(max, x);
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}
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while (pq.size() === n) {
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const [min, j, i] = pq.dequeue().element;
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if (max - min < r - l) {
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[l, r] = [min, max];
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}
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const iNext = i + 1;
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if (iNext < nums[j].length) {
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const next = nums[j][iNext];
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pq.enqueue([next, j, iNext]);
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max = Math.max(max, next);
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}
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}
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return [l, r];
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};
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const smallestRange = (nums: number[][]): number[] => {
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const pq = new MinPriorityQueue<[number, number, number]>({ priority: ([x]) => x });
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const n = nums.length;
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let [l, r, max] = [0, Number.POSITIVE_INFINITY, Number.NEGATIVE_INFINITY];
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for (let j = 0; j < n; j++) {
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const x = nums[j][0];
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pq.enqueue([x, j, 0]);
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max = Math.max(max, x);
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}
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while (pq.size() === n) {
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const [min, j, i] = pq.dequeue().element;
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if (max - min < r - l) {
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[l, r] = [min, max];
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}
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const iNext = i + 1;
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if (iNext < nums[j].length) {
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const next = nums[j][iNext];
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pq.enqueue([next, j, iNext]);
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max = Math.max(max, next);
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}
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}
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return [l, r];
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};

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