From ac82e7d7f550f51cb76324d3910436ec4c649c75 Mon Sep 17 00:00:00 2001
From: sanwuhong <shanruhongx@163.com>
Date: Mon, 6 Feb 2023 00:36:37 +0800
Subject: [PATCH] update algorithm

---
 .../java/com/learning/algorithm/ALGORITHM.md  | 33 +++++++++-
 .../com/learning/algorithm/Algorithm4.java    | 65 +++++++++++++++++++
 2 files changed, 97 insertions(+), 1 deletion(-)
 create mode 100644 src/main/java/com/learning/algorithm/Algorithm4.java

diff --git a/src/main/java/com/learning/algorithm/ALGORITHM.md b/src/main/java/com/learning/algorithm/ALGORITHM.md
index ebbf8fa..679507c 100644
--- a/src/main/java/com/learning/algorithm/ALGORITHM.md
+++ b/src/main/java/com/learning/algorithm/ALGORITHM.md
@@ -1233,6 +1233,28 @@ BFS与DFS相关的问题，经常都可以用两种方式求解，因此把相
 - [可能的二分法](https://leetcode.cn/problems/possible-bipartition/)
 
 ## 递归
+编写递归代码时最重要的有以下三点：
+1. 递归总有一个最简单的情况  — 方法的第一条语句总是一个包含return的条件语句。
+2. 递归调用总是去**尝试解决一个规模更小的子问题**，这样递归才能收敛到最简单的情况。如下代码第三、四参数的差值一直在缩小。
+3. 递归调用的父问题和尝试解决的子问题之间不应该有交集。
+
+```java
+class Solution {
+    public static int rank(int key, int[] a) {
+        return rank(key, a, 0, a.length - 1);
+    }
+
+    public static int rank(int key, int[] a ,int lo, int hi){
+        // 如果key存在于数组中，它的索引不会小于lo且不会大于hi
+        if(lo>hi) return -1;
+        int mid = lo + (hi -lo)/2;
+        if (key<a[mid]) return rank(key, a, lo, mid -1);
+        else if(key>a[mid]) return rank(key, a, mid+1, hi);
+        else return mid; 
+    }
+}
+```
+
 
 ### 递归三要素
 
@@ -2211,4 +2233,13 @@ public class Solution{
         
     }
 }
-```
\ No newline at end of file
+```
+
+# 算法4原书
+
+
+```java
+
+
+```
+
diff --git a/src/main/java/com/learning/algorithm/Algorithm4.java b/src/main/java/com/learning/algorithm/Algorithm4.java
new file mode 100644
index 0000000..31faf8b
--- /dev/null
+++ b/src/main/java/com/learning/algorithm/Algorithm4.java
@@ -0,0 +1,65 @@
+package com.learning.algorithm;
+
+public class Algorithm4 {
+
+    private double forMaxElement(double[] num) {
+        double max = num[0];
+        for (int i = 1; i < num.length; i++) {
+            if (num[i] > max) {
+                max = num[i];
+            }
+        }
+        return max;
+    }
+
+    public static boolean isPrime(int N) {
+        if (N < 2) {
+            return false;
+        }
+        for (int i = 2; i * i <= N; i++) {
+            if (N % i == 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+
+    /**
+     * 计算平方根
+     * 牛顿迭代法
+     *
+     * @param c
+     * @return
+     */
+    public static double sqrt(double c) {
+        if (c < 0) {
+            return Double.NaN;
+        }
+        double err = 1e-15;
+        double t = c;
+
+        while (Math.abs(t - c / t) > err * t) {
+            t = (c / t + t) / 2.0;
+        }
+        return t;
+    }
+
+    /**
+     * 计算调和级数
+     *
+     * @param N
+     * @return
+     */
+    public static double H(int N) {
+        double sum = 0.0;
+        for (int i = 0; i <= N; i++) {
+            sum += 1.0 / i;
+        }
+        return sum;
+    }
+
+    public static void main(String[] args) {
+        double sum = 0.0;
+        int cnt = 0;
+    }
+}