Merge pull request #1772 from symdunstaz/master

更新 背包理论基础01背包-1.md 中 java版本

problems/背包理论基础01背包-1.md

@@ -271,39 +271,64 @@ int main() {
 ### java 
 
 ```java
+public class BagProblem {
     public static void main(String[] args) {
-        int[] weight = {1, 3, 4};
+        int[] weight = {1,3,4};
-        int[] value = {15, 20, 30};
+        int[] value = {15,20,30};
-        int bagsize = 4;
+        int bagSize = 4;
-        testweightbagproblem(weight, value, bagsize);
+        testWeightBagProblem(weight,value,bagSize);
     }
 
-    public static void testweightbagproblem(int[] weight, int[] value, int bagsize){
+    /**
-        int wlen = weight.length, value0 = 0;
+     * 动态规划获得结果
-        //定义dp数组：dp[i][j]表示背包容量为j时，前i个物品能获得的最大价值
+     * @param weight  物品的重量
-        int[][] dp = new int[wlen + 1][bagsize + 1];
+     * @param value   物品的价值
-        //初始化：背包容量为0时，能获得的价值都为0
+     * @param bagSize 背包的容量
-        for (int i = 0; i <= wlen; i++){
+     */
-            dp[i][0] = value0;
+    public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
+
+        // 创建dp数组
+        int goods = weight.length;  // 获取物品的数量
+        int[][] dp = new int[goods][bagSize + 1];
+
+        // 初始化dp数组
+        // 创建数组后，其中默认的值就是0
+        for (int j = weight[0]; j <= bagSize; j++) {
+            dp[0][j] = value[0];
         }
-        //遍历顺序：先遍历物品，再遍历背包容量
+
-        for (int i = 1; i <= wlen; i++){
+        // 填充dp数组
-            for (int j = 1; j <= bagsize; j++){
+        for (int i = 1; i < weight.length; i++) {
-                if (j < weight[i - 1]){
+            for (int j = 1; j <= bagSize; j++) {
-                    dp[i][j] = dp[i - 1][j];
+                if (j < weight[i]) {
-                }else{
+                    /**
-                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
+                     * 当前背包的容量都没有当前物品i大的时候，是不放物品i的
+                     * 那么前i-1个物品能放下的最大价值就是当前情况的最大价值
+                     */
+                    dp[i][j] = dp[i-1][j];
+                } else {
+                    /**
+                     * 当前背包的容量可以放下物品i
+                     * 那么此时分两种情况：
+                     *    1、不放物品i
+                     *    2、放物品i
+                     * 比较这两种情况下，哪种背包中物品的最大价值最大
+                     */
+                    dp[i][j] = Math.max(dp[i-1][j] , dp[i-1][j-weight[i]] + value[i]);
                 }
             }
         }
-        //打印dp数组
+
-        for (int i = 0; i <= wlen; i++){
+        // 打印dp数组
-            for (int j = 0; j <= bagsize; j++){
+        for (int i = 0; i < goods; i++) {
-                System.out.print(dp[i][j] + " ");
+            for (int j = 0; j <= bagSize; j++) {
+                System.out.print(dp[i][j] + "\t");
             }
-            System.out.print("\n");
+            System.out.println("\n");
         }
     }
+}
+
 ```
 
 ### python