20181230

Author: mJackie

code/lc11.java

@@ -4,7 +4,7 @@
  * 题意：数组下标代表横坐标，数组中的值代表纵坐标，求最大面积
  * 难度：Medium
  * 分类：Array, Two Pointers
- * Tips：复杂度可以为O(N), 指针往里走, 若值也小了，则面积一定不会增大
+ * Tips：复杂度可以为O(N), 指针往里走, 若值也小了，则面积一定不会增大。和lc42做比较
  */
 public class lc11 {
     public static void main(String[] args) {

code/lc42.java

@@ -1,4 +1,7 @@
 package code;
+
+import java.util.Stack;
+
 /*
  * 42. Trapping Rain Water
  * 题意：能盛多少水
@@ -11,6 +14,7 @@ public class lc42 {
     public static void main(String[] args) {
         int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
         System.out.println(trap(height));
+        System.out.println(trap2(height));
     }
     public static int trap(int[] height) {
         if(height.length<3)
@@ -40,4 +44,23 @@ public static int trap(int[] height) {
         }
         return res;
     }
+
+    public static int trap2(int[] A) {
+        //栈方法
+        if (A==null) return 0;
+        Stack<Integer> s = new Stack<Integer>();
+        int i = 0, maxWater = 0, maxBotWater = 0;
+        while (i < A.length){
+            if (s.isEmpty() || A[i]<=A[s.peek()]){
+                s.push(i++);
+            }
+            else {
+                int bot = s.pop();
+                maxBotWater = s.isEmpty()? // empty means no il
+                        0:(Math.min(A[s.peek()],A[i])-A[bot])*(i-s.peek()-1);
+                maxWater += maxBotWater;
+            }
+        }
+        return maxWater;
+    }
 }

code/lc79.java

@@ -0,0 +1,42 @@
+package code;
+/*
+ * 79. Word Search
+ * 题意：在字符数组中搜索字符串
+ * 难度：Medium
+ * 分类：Array, Backtracking
+ * 思路：回溯法
+ * Tips：访问过的格子要标记，不能重复访问。回溯法注意回来的时候要重置标志位。向下找的时候直接找4个方向的，回来的时候不用再找了，只需重置标志位。
+ */
+public class lc79 {
+    public static void main(String[] args) {
+        char[][] board = {{'A','B','C','E'},{'S','F','C','S'},{'A','D','E','E'}};
+        System.out.println(exist(board,"SEE"));
+    }
+
+    public static boolean exist(char[][] board, String word) {
+        if(board.length==0)
+            return false;
+        boolean [][] flag = new boolean[board.length][board[0].length];
+        for (int i = 0; i < board.length ; i++) {
+            for (int j = 0; j < board[0].length ; j++) {
+                if(search(board,word,i,j,0,flag))
+                    return true;
+            }
+        }
+        return false;
+    }
+
+    public static boolean search(char[][] board, String word, int i, int j, int sum, boolean[][] flag){
+        if(sum==word.length())
+            return true;
+        if(i<0||j<0||i>=board.length||j>=board[0].length)
+            return false;
+        if(board[i][j]!=word.charAt(sum)||flag[i][j]==true)
+            return false;
+        flag[i][j] = true;
+        sum++;
+        boolean r = search(board, word, i, j+1, sum, flag)  || search(board, word, i, j-1, sum, flag) || search(board, word, i+1, j, sum, flag) ||search(board, word, i-1, j, sum, flag);
+        flag[i][j] = false;
+        return r;
+    }
+}

code/lc84.java

@@ -0,0 +1,66 @@
+package code;
+
+import java.util.Stack;
+
+/*
+ * 84. Largest Rectangle in Histogram
+ * 题意：直方图中最大矩形面积
+ * 难度：Hard
+ * 分类：Array, Stack
+ * 思路：两种方法：1.用dp找到边界，再遍历一遍; 2.用栈，栈内存索引，保证栈内索引对应的高度是递增的，若减了即找到了右边界，出栈开始计算。因为栈内是递增的，左边界就是上个栈内的元素。若栈为空，左边界就是-1。
+ * Tips：和lc42做比较，都可以用栈或者dp来做. 很难，栈的操作很难想到.
+ *
+ */
+public class lc84 {
+    public static void main(String[] args) {
+        int[] heights = {2,1,5,6,2,3};
+        System.out.println(largestRectangleArea(heights));
+        System.out.println(largestRectangleArea2(heights));
+    }
+
+    public static int largestRectangleArea(int[] heights) {
+        Stack<Integer> st = new Stack();
+        int res = 0;
+        for (int i = 0; i <= heights.length ; i++) {
+            int h = (i == heights.length ? 0 : heights[i]);
+            if( st.size()==0 || h>heights[st.peek()] ){ //递增入栈，保证栈内索引对应的Height递增
+                st.push(i);
+            }else{
+                int n = st.pop();
+                int left;
+                if(st.isEmpty())
+                    left = -1;  //若为空，则到最左边
+                else
+                    left = st.peek();
+                res = Math.max(res, heights[n] * (i-left-1)); //i之前的，要-1
+                i--;
+            }
+        }
+        return res;
+    }
+
+    public static int largestRectangleArea2(int[] heights) {
+        int[] leftMax = new int[heights.length];
+        int[] rightMax = new int[heights.length];
+        for (int i = 0; i < heights.length ; i++) { //get leftMax  注意这样dp时数组中保存的边界必须是i-1或i+1，否则无法dp传递
+            int p = i-1;
+            while( p>=0 && heights[p]>=heights[i]){
+                p = leftMax[p];
+            }
+            leftMax[i] = p;
+        }
+        for (int i = heights.length-1; i >= 0 ; i--) { //get rightMax
+            rightMax[i] = i;
+            int p = i+1;
+            while( p<heights.length && heights[p]>=heights[i]){
+                p = rightMax[p];
+            }
+            rightMax[i] = p;
+        }
+        int res = 0;
+        for (int i = 0; i < heights.length ; i++) {
+            res = Math.max(res,(rightMax[i]-leftMax[i]-1)*heights[i]);
+        }
+        return res;
+    }
+}

code/lc85.java

@@ -0,0 +1,49 @@
+package code;
+
+import java.util.Stack;
+
+/*
+ * 85. Maximal Rectangle
+ * 题意：为1的组成的最大矩形面积
+ * 难度：Hard
+ * 分类：Array, Hash Table, Dynamic Programming, Stack
+ * 思路：将问题转化为 84.Largest Rectangle in Histogram
+ * Tips：很难，lc84就够难了，没写过的谁能想到。。。
+ */
+public class lc85 {
+    public static void main(String[] args) {
+        char[][] matrix = {{'1','0','1','0','0'},
+                          {'1','0','1','1','1'},
+                          {'1','1','1','1','1'},
+                          {'1','0','0','1','0'}};
+        System.out.println(maximalRectangle(matrix));
+    }
+    public static int maximalRectangle(char[][] matrix) {
+        if(matrix.length==0||matrix[0].length==0)
+            return 0;
+        int col = matrix[0].length;
+        int[] row = new int[col];   // 用来记录直方图高度
+        int res =0;
+        for (int i = 0; i < matrix.length; i++) {
+            for (int j = 0; j < col ; j++) {
+                if(matrix[i][j]=='0')
+                    row[j] =0;
+                else
+                    row[j] = row[j]+1;
+            }
+            // 求直方图最大面积
+            Stack<Integer> st = new Stack();
+            for (int j = 0; j <= col ; j++) {
+                int h = ( j==col ? 0 : row[j] );
+                if(st.isEmpty() || h>row[st.peek()]){
+                    st.add(j);
+                }else{
+                    int index = st.pop();
+                    res = Math.max(res, row[index]*(j- ( st.isEmpty() ? -1 : st.peek() ) -1));
+                    j--;
+                }
+            }
+        }
+        return res;
+    }
+}