#业务算法# 求一个列表的C(m,n)的所有组合

data-structure-and-algorithms/README.md

@@ -1,3 +1,4 @@
 # 列表
   - [1~99，找重复数](https://github.com/liaojiacan/code-snippets/blob/master/data-structure-and-algorithms/src/main/java/com/github/liaojiacan/search/FindDuplicateNum.java)
   - [判断一个数是否是2的N次方](https://github.com/liaojiacan/code-snippets/blob/master/data-structure-and-algorithms/src/main/java/com/github/liaojiacan/LittleAlgorithms/CheckIsNthPowerOf2.java)
+  - [对一个列表求C(m,n)的所有组合数组](https://github.com/liaojiacan/code-snippets/blob/master/data-structure-and-algorithms/src/main/java/com/github/liaojiacan/combiner/Combiner.java)

data-structure-and-algorithms/src/main/java/com/github/liaojiacan/combine/Combiner.java

@@ -0,0 +1,65 @@
+package com.github.liaojiacan.combine;
+
+import com.sun.tools.javac.util.Assert;
+
+import java.util.ArrayList;
+import java.util.List;
+
+/**
+ * 列表的组合
+ * @see <a href="https://leetcode-cn.com/problems/combinations/submissions/">leetcode-Combinations</a>
+ * @author liaojiacan
+ * @date 2019/1/9
+ */
+public class Combiner {
+
+	/**
+	 * 获取originList的 大小为combinationSize的T元素的所有组合
+	 * @param originList
+	 * @param combinationSize
+	 * @param <T>
+	 * @return
+	 */
+	public static <T> List<List<T>> combine(List<T> originList, int combinationSize){
+		Assert.check(originList.size() >=combinationSize,"originList size should greater than combinationSize");
+		if(originList.size() == combinationSize ){
+			List<List<T>> result  = new ArrayList<>(1);
+			result.add(originList);
+			return result;
+		}
+		List<List<T>> combinations = new ArrayList<>(calculateCombinationNum(originList.size(),combinationSize));
+		combine(combinations,originList,combinationSize,0,new ArrayList<>(combinationSize));
+		return combinations;
+	}
+
+
+	private static <T> void combine(List<List<T>> combinations,List<T> originList,int combinationSize,int elementIndex,List<T> combination){
+
+		if( combinationSize == 0 ){
+			combinations.add(new ArrayList<>(combination));
+			return;
+		}
+
+		for( int i = elementIndex ; i <= originList.size()-combinationSize ; i++ ){
+			combination.add(originList.get(i));
+			combine(combinations,originList,combinationSize-1,i+1,combination);
+			combination.remove(combination.size()-1);
+		}
+
+	}
+
+	/**
+	 * 计算组合的近视值 C(m,n)
+	 * @see <a href="https://program-dog.blogspot.com/2017/05/stirlingapproximation.html">stirling</a>
+	 * @see <a href="https://blog.csdn.net/uself/article/details/54575930">组合的近似值算法</a>
+	 * @param originElementSize
+	 * @param combinationSize
+	 * @return
+	 */
+	private static int calculateCombinationNum(int originElementSize,int combinationSize){
+		int m = originElementSize;
+		int n = combinationSize;
+		return (int) Math.ceil((1/Math.sqrt(2*Math.PI))*Math.sqrt(m/(n*(m-n)))*(Math.pow(m,m)/Math.pow(n,n))*Math.pow((m-n),(n-m)));
+	}
+
+}