20190220

Author: mJackie

code/lc130.java

@@ -0,0 +1,54 @@
+package code;
+/*
+ * 130. Surrounded Regions
+ * 题意：把被 X 包围的 O 填补成 X
+ * 难度：Medium
+ * 分类：Depth-first Search, Breadth-first Search
+ * 思路：把四条外围边框的O向内延伸填为1，剩下的O全部替换为X，再把1替换成O
+ * Tips：
+ */
+public class lc130 {
+    public static void main(String[] args) {
+        char[][] board = {{'X','X','X','X'},{'X','O','O','X'},{'X','X','O','X'},{'X','O','X','X'}};
+        solve(board);
+        System.out.println(board);
+    }
+    public static void solve(char[][] board) {
+        if(board.length==0) return;
+        int[] row = {0, board.length-1};
+        for (int i = 0; i <row.length ; i++) {
+            for (int j = 0; j < board[0].length ; j++) {
+               if(board[row[i]][j]=='O'){     // 注意是row[i]，不是i
+                    dfs(board, row[i], j);
+                }
+            }
+        }
+        int[] col = {0, board[0].length-1};
+        for (int i = 0; i <col.length ; i++) {
+            for (int j = 0; j < board.length ; j++) {
+                if(board[j][col[i]]=='O'){
+                    dfs(board, j, col[i]);
+                }
+            }
+        }
+        for (int i = 0; i < board.length ; i++) {
+            for (int j = 0; j < board[0].length ; j++) {
+                if(board[i][j]=='1')
+                    board[i][j] = 'O';
+                else if(board[i][j]=='O')
+                    board[i][j] = 'X';
+            }
+        }
+    }
+
+    public static void dfs(char[][] board, int row, int col){
+        if(row<0||row>=board.length||col<0||col>=board[0].length) return;
+        if(board[row][col]=='O') {  //是 O 的话才继续递归
+            board[row][col] = '1';
+            dfs(board, row + 1, col);
+            dfs(board, row - 1, col);
+            dfs(board, row, col + 1);
+            dfs(board, row, col - 1);
+        }
+    }
+}

code/lc131.java

@@ -0,0 +1,49 @@
+package code;
+
+import java.util.ArrayList;
+import java.util.List;
+/*
+ * 131. Palindrome Partitioning
+ * 题意：回文切割，找出所有切法
+ * 难度：Medium
+ * 分类：Backtracking
+ * 思路：典型回溯法，注意向res添加内容时要重新new一下
+ * Tips：lc39
+ */
+public class lc131 {
+    public static void main(String[] args) {
+        List res = partition("aab");
+        System.out.println();
+    }
+    public static  List<List<String>> partition(String s) {
+        List<List<String>> res = new ArrayList<>();
+        helper(s, res, new ArrayList<>());
+        return res;
+    }
+
+    public static void helper(String s, List<List<String>> res, List<String> curr){
+        if(s.length()==0) {
+            res.add(new ArrayList<>(curr));  //新new一个
+        }
+        for (int i = 1; i <= s.length() ; i++) {
+            String curr_str = s.substring(0,i);
+            if(isPalindrome(curr_str)){
+                curr.add(curr_str);
+                helper(s.substring(i), res, curr);
+                curr.remove(curr.size()-1);     //不要用curr.remove(curr_str); 会把集合内顺序打乱，AC不了
+            }
+        }
+    }
+
+    public static boolean isPalindrome(String s){   //判断是否为回文
+        int b =0;
+        int e = s.length()-1;
+        while(b<e){
+            if(s.charAt(b)!=s.charAt(e))
+                return false;
+            b++;
+            e--;
+        }
+        return true;
+    }
+}

code/lc134.java

@@ -0,0 +1,27 @@
+package code;
+/*
+ * 134. Gas Station
+ * 题意：给两个数组，一个表示在该点能加多少油，一个表示该点到下个点消耗多少油，加油站连成一个环，问能否走遍所有加油站，从哪一站开始
+ * 难度：Medium
+ * 分类：Greedy
+ * 思路：计算两个数组的差，再把每个位置加起来，如果>=0，则能跑完。位置为最大连续子数组的开始位置。
+ * Tips：lc53
+ */
+public class lc134 {
+    public int canCompleteCircuit(int[] gas, int[] cost) {
+        int[] arr = new int[gas.length];
+        int sum = 0, pos = 0, dp =0;
+        for (int i = 0; i < arr.length ; i++) {
+            arr[i] = gas[i] - cost[i];
+            sum += arr[i];
+
+            if(dp<=0) { //最大连续子数组，并记录位置
+                dp = arr[i];
+                pos = i;
+            }
+            else
+                dp += arr[i];
+        }
+        return sum<0? -1: pos;
+    }
+}

code/lc140.java

@@ -0,0 +1,39 @@
+package code;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+import java.util.List;
+/*
+ * 140. Word Break II
+ * 题意：给定词典，比照词典切字符串，能切出几种情况
+ * 难度：Hard
+ * 分类：Dynamic Programming, Backtracking
+ * 思路：回溯法
+ * Tips：和139做对比，因为需要每步骤返回值，所以递归的方法在这题更合适
+ */
+public class lc140 {
+    public List<String> wordBreak(String s, List<String> wordDict) {
+        List<String> res = new ArrayList<>();
+        if(s.length()==0) return res;
+        return helper(s, wordDict, new HashMap<>());
+    }
+
+    public List<String> helper(String s, List<String> wordDict, HashMap<String, List<String>> hm){
+        List<String> res = new ArrayList<>();
+        if(hm.containsKey(s)) return hm.get(s);     //hm用以记录子结果，之前递归已计算出，则直接返回。备忘录算法。
+        for (int i = s.length()-1; i >=0 ; i--) {
+            String cut_str = s.substring(i);
+            if(wordDict.contains(cut_str)){
+                List<String> ls = helper(s.substring(0,i), wordDict, hm);   //切掉以后剩下的字符串的结果
+                for(String str:ls){
+                    res.add(str+" "+cut_str);   //结果拼上切掉的字符串
+                }
+                if(i==0){   //切完了，不需要拼空格。hm中没有保存键值为""的，上个循环不会有效果。
+                    res.add(cut_str);
+                }
+                hm.put(s, res);
+            }
+        }
+        return res;
+    }
+}

readme.md

@@ -88,6 +88,9 @@ Language: Java
 | 125 [Java](./code/lc125.java)
 | 127 [Java](./code/lc127.java)
 | 128 [Java](./code/lc128.java)
+| 130 [Java](./code/lc130.java)
+| 131 [Java](./code/lc131.java)
+| 134 [Java](./code/lc134.java)
 | 136 [Java](./code/lc136.java)
 | 139 [Java](./code/lc139.java)
 | 141 [Java](./code/lc141.java)