20190312

code/lc120.java

+package code;
+
+import java.util.List;
+/*
+ * 120. Triangle
+ * 题意：三角形的矩阵，求值最小的路径
+ * 难度：Medium
+ * 分类：Array, Dynamic Porgramming
+ * 思路：动态规划 dp[i] = min(dp[i-1],dp[i]) + val
+ * Tips：
+ */
+public class lc120 {
+    public int minimumTotal(List<List<Integer>> triangle) {
+        int len = triangle.get(triangle.size()-1).size();
+        int[] dp = new int[len];
+        int[] dp2 = new int[len];
+        int index = 0;
+        int res = Integer.MAX_VALUE;
+        while(index<triangle.size()){
+            List<Integer> ls = triangle.get(index);
+            res = Integer.MAX_VALUE;
+            for (int i = 0; i < ls.size() ; i++) {
+                if(i==0) dp2[i] = dp[i] + ls.get(i);
+                else if(i==ls.size()-1) dp2[i] = dp[i-1] + ls.get(i);   //注意是 i==ls.size()-1
+                else dp2[i] = Math.min(dp[i-1],dp[i]) + ls.get(i);
+                res = Math.min(dp2[i], res);
+            }
+            dp = dp2;   //两个一维数组
+            dp2 = new int[len];
+            index++;
+        }
+        return res;
+    }
+}

code/lc213.java

+package code;
+
+import java.util.Arrays;
+/*
+ * 213. House Robber II
+ * 题意：数组最大和，不能选取相邻的两个数。数组首尾是连着的
+ * 难度：Medium
+ * 分类：Dynamic Programming
+ * 思路：分别把第一个元素置0和最后一个元素置0，用lc198的解法
+ *
+ * Tips：lc198
+ */
+public class lc213 {
+    public int rob(int[] nums) { //分别把第一个元素置0和最后一个元素置0，用lc198的解法
+        if(nums.length == 0)
+            return 0;
+        if(nums.length == 1)
+            return nums[0];
+        int res1 = helper(Arrays.copyOf(nums, nums.length-1));
+        nums[0] = 0;
+        int res2 = helper(nums);
+        return Math.max(res1, res2);
+    }
+
+    public int helper(int[] nums) {
+        if(nums.length == 0)
+            return 0;
+        if(nums.length == 1)
+            return nums[0];
+        int[] dp = new int[nums.length];
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for (int i = 2; i < nums.length ; i++) {
+            dp[i] = Math.max((dp[i-2] + nums[i]),dp[i-1]);  //dp[i] 表示以 0~i 的数组的结果
+        }
+        return dp[nums.length-1];
+    }
+}