Merge pull request algorithm004-04#1026 from zenglinhui/master

044-Week 07

Week 07/id_044/LeetCode_044_146.java

@@ -0,0 +1,32 @@
+package week_07.lesson17;
+
+import java.util.LinkedHashMap;
+import java.util.Map;
+
+public class LRUCache extends LinkedHashMap<Integer, Integer> {
+
+    private int capacity;
+
+    public LRUCache(int capacity) {
+        super(capacity,0.75F,true);
+        this.capacity = capacity;
+    }
+
+    public int get(int key) {
+        return super.getOrDefault(key, -1);
+    }
+
+    public void put(int key, int value) {
+        super.put(key,value);
+    }
+
+    @Override
+    protected boolean removeEldestEntry(Map.Entry<Integer, Integer> eldest) {
+        return size() > capacity;
+    }
+
+    public static void main(String[] args) {
+
+    }
+
+}

Week 07/id_044/LeetCode_044_190.java

@@ -0,0 +1,23 @@
+package week_07.lesson16;
+
+public class ReverseBits {
+
+    public int reverseBits(int n) {
+        int result = 0;
+        for (int i = 0; i < 32; i++) {
+            //右移i位
+            int tmp = n >> i;
+            //有效位是否为1
+            tmp = tmp & 1;
+            //反转位置
+            tmp = tmp << (31 - i);
+            //
+            result = result | tmp;
+        }
+        return result;
+    }
+
+    public static void main(String[] args) {
+
+    }
+}

Week 07/id_044/LeetCode_044_191.java

@@ -0,0 +1,30 @@
+package week_07.lesson16;
+
+public class NumberOfOneBits {
+
+    public int hammingWeight(int n) {
+        /*int bits = 0;
+        //循环加标记法，每次把标记左移一位，然后&上n，是否为1
+        int mask = 1;
+        for (int i = 0; i < 32; i++) {
+            if (0 != (n & mask)) {
+                bits++;
+            }
+            mask = mask << 1;
+        }
+        return bits;*/
+
+        //用n&(n-1)来运算，在二进制表示中，总是会把最低位(也可理解成最后一位)的1变成0
+        int sum = 0;
+        while (n != 0) {
+            sum++;
+            n = n & (n - 1);
+        }
+        return sum;
+    }
+
+    public static void main(String[] args) {
+
+    }
+
+}

Week 07/id_044/LeetCode_044_231.java

@@ -0,0 +1,26 @@
+package week_07.lesson16;
+
+public class PowerOfTwo {
+
+    /**
+     * 若 n = 2^x 且 x 为自然数（即 n 为 2 的幂）,则一定满足以下条件：
+     * 1.恒有 n & (n - 1) == 0,这是因为：
+     *      n 二进制最高位为1，其余所有位为0；
+     *      2^0     --> 0001
+     *      2^1     --> 0010
+     *      2^2     --> 0100
+     *      2^3     --> 1000
+     *      n - 1 二进制最高位为0，其余所有位为1；
+     * 2.一定满足 n > 0
+     * @param n
+     * @return
+     */
+    public boolean isPowerOfTwo(int n) {
+        return n > 0 && (n & (n - 1)) == 0;
+    }
+
+    public static void main(String[] args) {
+
+    }
+
+}

Week 07/id_044/LeetCode_044_242.java

@@ -0,0 +1,45 @@
+package week_07.lesson18;
+
+import java.util.Arrays;
+
+public class ValidAnagram {
+
+
+    public static boolean isAnagram(String s, String t) {
+
+        /*if (s.length() != t.length()) {
+            return false;
+        }
+
+        int[] counter = new int[26];
+        int[] diffcount = new int[26];
+        for (int i = 0; i < s.length(); i++) {
+            counter[s.charAt(i) - 'a']++;
+            counter[t.charAt(i) - 'a']--;
+        }
+
+        return Arrays.equals(counter,diffcount);*/
+
+        if (s.length() != t.length()) {
+            return false;
+        }
+
+        int[] counter = new int[26];
+        for (int i = 0; i < s.length(); i++) {
+            counter[s.charAt(i) - 'a']++;
+            counter[t.charAt(i) - 'a']--;
+        }
+        for (int i = 0; i < s.length(); i++) {
+            if (counter[s.charAt(i) - 'a'] != 0) {
+                return false;
+            }
+        }
+        return true;
+
+    }
+
+    public static void main(String[] args) {
+
+    }
+
+}

Week 07/id_044/NOTE.md

@@ -1,4 +1,57 @@
 # NOTE
+//冒泡排序
+    private static void bubbleSort(int[] arr) {
+
+        int len = arr.length;
+        for (int i = 0; i < len - 1; i++) {
+            for (int j = 0; j < len - 1 - i; j++) {
+                if (arr[j] > arr[j + 1]) {
+                    int temp = arr[j + 1];
+                    arr[j + 1] = arr[j];
+                    arr[j] = temp;
+                }
+            }
+        }
+
+    }
+
+//选择排序
+    private static void selectionSort(int[] arr) {
+        int len = arr.length;
+        int minIndex,temp;
+        for (int i = 0; i < len - 1; i++) {
+            minIndex = i;
+            for (int j = i + 1; j < len; j++) {
+                if (arr[j] < arr[minIndex]) {
+                    minIndex = j;
+                }
+            }
+            temp = arr[i];
+            arr[i] = arr[minIndex];
+            arr[minIndex] = temp;
+        }
+    }
+
+//插入排序
+    private static void insertionSort(int[] arr) {
+        int len = arr.length;
+        int preIndex,current;
+        for (int i = 1; i < len; i++) {
+            preIndex = i - 1;
+            current = arr[i];
+            //从preIndex位置开始，向前查找，如果前一个元素比后一个元素大，则交换
+            while (preIndex > -1 && arr[preIndex] > current) {
+                arr[preIndex + 1] = arr[preIndex];
+                preIndex--;
+            }
+            arr[preIndex + 1] = current;
+        }
+    }
+
+
+
+
+