Add solution #1030

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,101 LeetCode solutions in JavaScript
+# 1,102 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -837,6 +837,7 @@
 1027|[Longest Arithmetic Subsequence](./solutions/1027-longest-arithmetic-subsequence.js)|Medium|
 1028|[Recover a Tree From Preorder Traversal](./solutions/1028-recover-a-tree-from-preorder-traversal.js)|Hard|
 1029|[Two City Scheduling](./solutions/1029-two-city-scheduling.js)|Medium|
+1030|[Matrix Cells in Distance Order](./solutions/1030-matrix-cells-in-distance-order.js)|Easy|
 1037|[Valid Boomerang](./solutions/1037-valid-boomerang.js)|Easy|
 1038|[Binary Search Tree to Greater Sum Tree](./solutions/1038-binary-search-tree-to-greater-sum-tree.js)|Medium|
 1041|[Robot Bounded In Circle](./solutions/1041-robot-bounded-in-circle.js)|Medium|

solutions/1030-matrix-cells-in-distance-order.js

@@ -0,0 +1,34 @@
+/**
+ * 1030. Matrix Cells in Distance Order
+ * https://leetcode.com/problems/matrix-cells-in-distance-order/
+ * Difficulty: Easy
+ *
+ * You are given four integers row, cols, rCenter, and cCenter. There is a rows x cols matrix and
+ * you are on the cell with the coordinates (rCenter, cCenter).
+ *
+ * Return the coordinates of all cells in the matrix, sorted by their distance from (rCenter,
+ * cCenter) from the smallest distance to the largest distance. You may return the answer in any
+ * order that satisfies this condition.
+ *
+ * The distance between two cells (r1, c1) and (r2, c2) is |r1 - r2| + |c1 - c2|.
+ */
+
+/**
+ * @param {number} rows
+ * @param {number} cols
+ * @param {number} rCenter
+ * @param {number} cCenter
+ * @return {number[][]}
+ */
+var allCellsDistOrder = function(rows, cols, rCenter, cCenter) {
+  const result = [];
+  for (let row = 0; row < rows; row++) {
+    for (let col = 0; col < cols; col++) {
+      result.push([row, col]);
+    }
+  }
+  return result.sort((a, b) => {
+    return Math.abs(a[0] - rCenter) + Math.abs(a[1] - cCenter)
+      - Math.abs(b[0] - rCenter) - Math.abs(b[1] - cCenter);
+  });
+};