多数元素 and 轮转数组

com/zerroi/leetcode/ThreeFive/MajorityElement.java

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+package com.zerroi.leetcode.ThreeFive;
+
+import java.util.Arrays;
+import java.util.HashMap;
+import java.util.Map;
+
+public class MajorityElement {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        int[] nums = new int[]{2, 2, 1, 1, 1, 2, 2};
+        int res = solution.majorityElement(nums);
+//        System.out.println("res = " + res);
+//        System.out.println(5/2);
+    }
+}
+
+/**
+ * 给定一个大小为 n 的数组 nums ，返回其中的多数元素。多数元素是指在数组中出现次数 大于 ⌊ n/2 ⌋ 的元素。
+ * 你可以假设数组是非空的，并且给定的数组总是存在多数元素。
+ */
+class Solution {
+    public int majorityElement(int[] nums) {
+//        用HashMap key->元素 value->元素出现的次数
+        Map<Integer, Integer> map = new HashMap<>();
+        for (int num : nums) {
+            if (map.containsKey(num)) {
+                map.put(num, map.get(num) + 1);
+            } else {
+                map.put(num, 1);
+            }
+        }
+
+        for (Integer key : map.keySet()) {
+            if (map.get(key) > nums.length / 2) {
+                return key;
+            }
+        }
+        return 0;
+    }
+
+    public int majorityElementSort(int[] nums) {
+        Arrays.sort(nums);
+        return nums[nums.length / 2];
+    }
+
+    public int majorityElementVote(int[] nums) {
+        int winner = nums[0];
+        int count = 1;
+        for (int i = 1; i < nums.length; i++) {
+            if (nums[i] == winner) {
+                count++;
+            } else if (count == 0) {
+                winner = nums[i];
+                count++;
+            } else {
+                count--;
+            }
+        }
+        return winner;
+    }
+}

com/zerroi/leetcode/ThreeFive/Rotate.java

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+package com.zerroi.leetcode.ThreeFive;
+
+import java.util.Arrays;
+
+/**
+ * 给定一个整数数组 nums，将数组中的元素向右轮转 k 个位置，其中 k 是非负数。
+ */
+public class Rotate {
+    public static void main(String[] args) {
+
+    }
+}
+
+class SolutionSecond {
+    public void rotate(int[] nums, int k) {
+        k = k % nums.length;
+        int[] temp = new int[k];
+        for (int i = 0; i < k; i++) {
+            temp[i] = nums[nums.length - k + i];
+        }
+
+        for (int i = nums.length - k; i >= 0; i--) {
+            nums[i + k] = nums[i];
+        }
+
+        for (int i = 0; i < k; i++) {
+            nums[i] = temp[i];
+        }
+    }
+
+    public void rotateReverse(int[] nums, int k) {
+        k = k % nums.length;
+        reverse(nums, 0, nums.length);
+        reverse(nums, 0, k - 1);
+        reverse(nums, k, nums.length);
+    }
+
+    public void reverse(int[] nums, int start, int end) {
+        while (start < end) {
+            int temp = nums[start];
+            nums[start] = nums[end];
+            nums[end] = temp;
+            start += 1;
+            end -= 1;
+        }
+    }
+}