LeetCode——第163場周賽

5263. 二維網格遷移

題目很簡單，看下面代碼

 public List<List<Integer>> shiftGrid(int[][] grid, int k) {
     List<List<Integer>> lists = new ArrayList<>();
     for (int i = 0; i < grid[0].length; i++) {
         List<Integer> list = new ArrayList<>();
         for (int j = 0; j < grid.length; j++) {
             list.add(grid[j][i]);
         }
         lists.add(list);
     }
     if (k == 0) {
         return lists;
     }
     for (int i = 0; i < k; i++) {
         List<Integer> list = lists.remove(lists.size() - 1);
         Integer m = list.remove(list.size() - 1);
         list.add(0, m);
         lists.add(0, list);
     }
     List<List<Integer>> lists2 = new ArrayList<>();
     for (int i = 0; i < lists.get(0).size(); i++) {
         List<Integer> list = new ArrayList<>();
         for (int j = 0; j < lists.size(); j++) {
             list.add(lists.get(j).get(i));
         }
         lists2.add(list);
     }
     return lists2;
 }

5264. 在受污染的二叉樹中查找元素

也比較容易，在構造方法裏面直接就用DFS把這個樹恢復好，用Set存一下遍歷過程中的值。

 class FindElements {

     HashSet<Integer> set = new HashSet<>();

     public FindElements(TreeNode root) {
         if (root == null) {
             return;
         }
         root.val = 0;
         DFS(root);
     }

     private void DFS(TreeNode root) {
         if (root == null) {
             return;
         }
         set.add(root.val);
         if (root.left != null) {
             root.left.val = 2 * root.val + 1;
         }
         if (root.right != null) {
             root.right.val = 2 * root.val + 2;
         }
         DFS(root.left);
         DFS(root.right);
     }

     public boolean find(int target) {
         return set.contains(target);
     }
 }

5265. 可被三整除的最大和

• 剛開始感覺像是dp，沒有啥思路。
• 回溯很簡單，但是肯定超時。
• 最後發現是個數學題。
• 將%3==1和%3==2的元素存起來，然後對這兩個列表排序。
• 判斷整個數組的sum%3==?，然後在對應的兩個取餘列表裏面減就可以了。

 public int maxSumDivThree(int[] nums) {
     List<Integer> list1 = new ArrayList<>(nums.length);
     List<Integer> list2 = new ArrayList<>();
     int sum = 0;
     for (int num : nums) {
         if (num % 3 == 2) {
             list2.add(num);
         } else if (num % 3 == 1) {
             list1.add(num);
         }
         sum += num;
     }
     Collections.sort(list1);
     Collections.sort(list2);
     if (sum % 3 == 0) {
         return sum;
     } else if (sum % 3 == 1) {
         int a1 = Integer.MAX_VALUE, a2 = Integer.MAX_VALUE;
         if (list1.size() > 0) {
             a1 = list1.get(0);
         }
         if (list2.size() > 1) {
             a2 = list2.get(0) + list2.get(1);
         }
         return sum - Math.min(a1, a2);
     } else {
         int a1 = Integer.MAX_VALUE, a2 = Integer.MAX_VALUE;
         if (list2.size() > 0) {
             a2 = list2.get(0);
         }
         if (list1.size() > 1) {
             a1 = list1.get(0) + list1.get(1);
         }
         return sum - Math.min(a1, a2);
     }
 }

5266. 推箱子

這個題我不會，我嘗試用DFS 暴力暴出來，但是代碼龐大且複雜。

我寫吐了，下面是有問題的代碼

 int[] T = new int[2];
 int[] B = new int[2];
 int[] S = new int[2];

 public int minPushBox(char[][] grid) {

     for (int i = 0; i < grid.length; i++) {
         for (int j = 0; j < grid[0].length; j++) {
             if (grid[i][j] == 'T') {
                 T[0] = i;
                 T[1] = j;
             } else if (grid[i][j] == 'B') {
                 B[0] = i;
                 B[1] = j;
             } else if (grid[i][j] == 'S') {
                 S[0] = i;
                 S[1] = j;
             }
         }
     }
     boolean[][] isVisit = new boolean[grid.length][grid[0].length];
     DFS(B[0], B[1], S[0], S[1], isVisit, grid);
     return min;
 }

 int tem = 0;
 int min = Integer.MAX_VALUE;

 private void DFS(int bi, int bj, int si, int sj, boolean[][] isVisit, char[][] grid) {
     if (bi < 0 || bi >= grid.length
             || bj < 0 || bj >= grid[0].length
             || grid[bi][bj] == '.'
             || isVisit[bi][bj]) {
         return;
     }
     if (bi == T[0] && bj == T[1]) {
         min = Math.min(tem, min);
         return;
     }
     isVisit[bi][bj] = true;
     if (canGo(bi, bj, bi + 1, bj, si, sj, grid, new boolean[grid.length][grid[0].length])) {
         tem++;
         DFS(bi - 1, bj, bi, bj, isVisit, grid);
         tem--;
     }
     if (canGo(bi, bj, bi - 1, bj, si, sj, grid, new boolean[grid.length][grid[0].length])) {
         tem++;
         DFS(bi + 1, bj, bi, bj, isVisit, grid);
         tem--;
     }
     if (canGo(bi, bj, bi, bj + 1, si, sj, grid, new boolean[grid.length][grid[0].length])) {
         tem++;
         DFS(bi, bj - 1, bi, bj, isVisit, grid);
         tem--;
     }
     if (canGo(bi, bj, bi, bj - 1, si, sj, grid, new boolean[grid.length][grid[0].length])) {
         tem++;
         DFS(bi, bj + 1, bi, bj, isVisit, grid);
         tem--;
     }
     isVisit[bi][bj] = false;
 }

 private boolean canGo(int bi, int bj, int ti, int tj, int si, int sj, char[][] grid, boolean[][] isVisit) {
     if (si < 0 || si >= grid.length
             || sj < 0 || sj >= grid[0].length
             || (grid[si][sj] == '.' || (si == bi && sj == bj))
             || isVisit[si][sj]) {
         return false;
     }
     isVisit[si][sj] = true;
     if (ti == si && tj == sj) {
         return true;
     }
     boolean end = canGo(bi, bj, ti, tj, si - 1, sj, grid, isVisit)
             || canGo(bi, bj, ti, tj, si - 1, sj, grid, isVisit)
             || canGo(bi, bj, ti, tj, si - 1, sj, grid, isVisit)
             || canGo(bi, bj, ti, tj, si - 1, sj, grid, isVisit);
     isVisit[si][sj] = false;
     return end;
 }