20190507

code/lc343.java

+package code;
+/*
+ * 343. Integer Break
+ * 题意：给定一个正整数，找出一组数字和为该数，且这组数的乘积最大
+ * 难度：Medium
+ * 分类：Math, Dynamic Programming
+ * 思路：能减3的减3，因为3得到的乘积最大
+ * Tips：
+ */
+public class lc343 {
+    public int integerBreak(int n) {
+        if(n==2) return 1;
+        if(n==3) return 2;
+        return helper(n);
+    }
+
+    public int helper(int n){
+        if(n==2) return 2;
+        if(n==3) return 3;
+        if(n==4) return 4;
+        if(n==5) return 6;
+        else return helper(n-3)*3;
+    }
+}

code/lc685.java

+package code;
+/*
+ * 685. Redundant Connection II
+ * 题意：一个图，删掉一个边，使其成为一个根树
+ * 难度：Hard
+ * 分类：Tree, Depth-first Search, Union Find, Graph
+ * 思路：要把问题想清楚
+ *      判断是否有某个节点父节点有两个, 记为e1, e2
+ *      再判断是否有环
+ *      4中情况，分别想清楚返回什么
+ *      自己没想清楚两种情况的交叉，以为判断完第一步就可直接返回
+ *      如何判断有环，可以利用并查集的思想
+ * Tips：https://leetcode.com/problems/redundant-connection-ii/discuss/108045/C%2B%2BJava-Union-Find-with-explanation-O(n)
+ */
+public class lc685 {
+    public int[] findRedundantDirectedConnection(int[][] edges) {
+        int[] can1 = {-1, -1};
+        int[] can2 = {-1, -1};
+        int[] parent = new int[edges.length + 1];
+        for (int i = 0; i < edges.length; i++) {
+            if (parent[edges[i][1]] == 0) {
+                parent[edges[i][1]] = edges[i][0];
+            } else {
+                can2 = new int[] {edges[i][0], edges[i][1]};
+                can1 = new int[] {parent[edges[i][1]], edges[i][1]};
+                edges[i][1] = 0;
+            }
+        }
+        for (int i = 0; i < edges.length; i++) {
+            parent[i] = i;
+        }
+        for (int i = 0; i < edges.length; i++) {
+            if (edges[i][1] == 0) {
+                continue;
+            }
+            int child = edges[i][1], father = edges[i][0];
+            if (root(parent, father) == child) {    //判断father的父节点是不是child
+                if (can1[0] == -1) {
+                    return edges[i];
+                }
+                return can1;
+            }
+            parent[child] = father;
+        }
+        return can2;
+    }
+
+    int root(int[] parent, int i) {
+        while (i != parent[i]) {    //找到根为止
+            parent[i] = parent[parent[i]];
+            i = parent[i];
+        }
+        return i;
+    }
+}

readme.md

@@ -190,7 +190,8 @@ LeetCode 指南
 | 337 [Java](./code/lc337.java)
 | 338 [Java](./code/lc338.java)
 | 341 [Java](./code/lc341.java)
-| 344 [Java](./code/lc338.java)
+| 343 [Java](./code/lc343.java)
+| 344 [Java](./code/lc344.java)
 | 347 [Java](./code/lc347.java)
 | 350 [Java](./code/lc350.java)
 | 371 [Java](./code/lc371.java)
@@ -217,6 +218,7 @@ LeetCode 指南
 | 617 [Java](./code/lc617.java)
 | 621 [Java](./code/lc621.java)
 | 647 [Java](./code/lc647.java)
+| 685 [Java](./code/lc685.java)
 | 714 [Java](./code/lc714.java)
 | 746 [Java](./code/lc746.java)
 | 771 [Java](./code/lc771.java)