Add solution #503

JoshCrozier · JoshCrozier · commit a2cd98b16900 · 2023-01-20T19:28:29.000-06:00
diff --git a/0503-next-greater-element-ii.js b/0503-next-greater-element-ii.js
@@ -0,0 +1,33 @@
+/**
+ * 503. Next Greater Element II
+ * https://leetcode.com/problems/next-greater-element-ii/
+ * Difficulty: Medium
+ *
+ * Given a circular integer array nums (i.e., the next element of
+ * nums[nums.length - 1] is nums[0]), return the next greater number
+ * for every element in nums.
+ *
+ * The next greater number of a number x is the first greater number
+ * to its traversing-order next in the array, which means you could
+ * search circularly to find its next greater number. If it doesn't
+ * exist, return -1 for this number.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var nextGreaterElements = function(nums) {
+  const total = nums.length;
+  const result = new Array(total).fill(-1);
+  const stack = [];
+
+  for (let i = 0; i < 2 * total; i++) {
+    while (stack.length && nums[i % total] > nums[stack[stack.length - 1]]) {
+      result[stack.pop()] = nums[i % total];
+    }
+    stack.push(i % total);
+  }
+
+  return result;
+};
diff --git a/README.md b/README.md
@@ -165,6 +165,7 @@
 500|[Keyboard Row](./0500-keyboard-row.js)|Easy|
 501|[Find Mode in Binary Search Tree](./0501-find-mode-in-binary-search-tree.js)|Easy|
 502|[IPO](./0502-ipo.js)|Hard|
+503|[Next Greater Element II](./0503-next-greater-element-ii.js)|Medium|
 506|[Relative Ranks](./0506-relative-ranks.js)|Easy|
 509|[Fibonacci Number](./0509-fibonacci-number.js)|Easy|
 520|[Detect Capital](./0520-detect-capital.js)|Easy|
