add leetcode: offer-39-66

Author: CodeWater404

Algorithm/src/offer/_39TheNumberOfMoreThanHalfOfTheArrayAppearsMoreThanHalf.java

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+package offer;
+
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * @author ： CodeWater
+ * @create ：2022-07-15-21:51
+ * @Function Description ：39.数组中出现次数超过一半的数字
+ */
+public class _39TheNumberOfMoreThanHalfOfTheArrayAppearsMoreThanHalf {
+    //==========================hashmap============================
+    class Solution {
+        public int majorityElement(int[] nums) {
+            Map<Integer , Integer> hash = new HashMap<>();
+            for( int num : nums ){
+                if( hash.containsKey(num) ) hash.put(num , hash.get(num) + 1 );
+                else hash.put( num , 1 );
+            }
+            for( Map.Entry<Integer , Integer> entry : hash.entrySet() ){
+                if( entry.getValue() > nums.length / 2 ) return entry.getKey();
+            }
+            return -1;
+        }
+    }
+    
+    //==========================摩尔投票法============================
+    class Solution2 {
+        public int majorityElement(int[] nums) {
+            // x假定是出现次数最多的众数   votes票数
+            int x = 0 , votes = 0;
+            for( int num : nums ){
+                // 票数为0是 假定当前num是众数
+                if(votes == 0 ) x = num;
+                // 判断num是不是上一轮选定的众数x  是的话票数+1 ， 不是-1
+                votes += num == x ? 1 : -1 ;
+            }
+            // 最终众数会赋给x ， 其余的都会在正负抵消
+            return x;
+        }
+    }
+}

Algorithm/src/offer/_66BuildAMultiplicationGroup.java

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+package offer;
+
+/**
+ * @author ： CodeWater
+ * @create ：2022-07-15-23:13
+ * @Function Description ：66.构建乘积数组
+ * 
+ */
+public class _66BuildAMultiplicationGroup {
+    class Solution {
+        // 题解,看图对比代码比较好理解https://leetcode.cn/problems/gou-jian-cheng-ji-shu-zu-lcof/solution/mian-shi-ti-66-gou-jian-cheng-ji-shu-zu-biao-ge-fe/
+        public int[] constructArr(int[] a) {
+            int len = a.length;
+            if( len == 0 ) return new int[0];
+            int[] b = new int[len];
+            b[0] = 1;
+
+            // 正向遍历求三角,从1开始，依次递增乘积
+            for( int i = 1 ; i < len ; i++ ){
+                b[i] = b[i - 1] * a[i - 1];
+            }
+            // 正向遍历上一个下标会储存前面的乘积，倒序遍历只能开一个temp临时存储
+            int temp = 1;
+            // 倒序遍历，从倒数第二个开始，依次递增乘积
+            for( int i = len - 2 ; i >= 0 ; i-- ){
+                temp *= a[i + 1];
+                b[i] *= temp;
+            }
+
+            return b;
+        }
+    }
+}