79:单词搜索
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src/main/java/leetcode/editor/cn/WordSearch.java
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src/main/java/leetcode/editor/cn/WordSearch.java
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//给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
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//
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// 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "AB
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//CCED"
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//输出：true
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//
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//
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// 示例 2：
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//
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//
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//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SE
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//E"
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//输出：true
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//
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//
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// 示例 3：
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//
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//
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//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "AB
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//CB"
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//输出：false
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//
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//
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//
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//
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// 提示：
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//
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//
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// m == board.length
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// n = board[i].length
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// 1 <= m, n <= 6
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// 1 <= word.length <= 15
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// board 和 word 仅由大小写英文字母组成
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//
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//
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//
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//
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// 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
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// Related Topics 数组 回溯 矩阵
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// 👍 973 👎 0
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package leetcode.editor.cn;
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import java.util.Arrays;
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//79:单词搜索
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public class WordSearch{
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public static void main(String[] args) {
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//测试代码
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Solution solution = new WordSearch().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public boolean exist(char[][] board, String word) {
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String str = Arrays.deepToString(board);
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for (int i = 0; i < word.length(); i++) {
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if (str.indexOf(word.charAt(i)) == -1) {
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return false;
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}
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}
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for (int i = 0; i < board.length; i++) {
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for (int j = 0; j < board[0].length; j++) {
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if (board[i][j] == word.charAt(0)) {
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if (dfs(board, word, i, j, 0, new int[board.length][board[0].length])) {
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return true;
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}
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}
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}
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}
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return false;
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}
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/**
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* 深度优先搜索
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* @param board 传入的二维字符网格
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* @param word 目标字符串
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* @param i 起点 i 坐标
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* @param j 起点 j 坐标
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* @param index 遍历到word的哪一个位置了
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* @param visit 访问数组，记录哪些地方已经走过
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* @return
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*/
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private boolean dfs(char[][] board, String word, int i, int j, int index, int[][] visit) {
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if (i < 0 || i >= board.length || j < 0 || j >= board[0].length
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|| board[i][j] != word.charAt(index) || visit[i][j] == 1) {
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return false;
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}
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if (index == word.length() - 1) {
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return true;
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}
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visit[i][j] = 1;
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int[][] direction = new int[][]{{i - 1, j}, {i + 1, j}, {i, j - 1}, {i, j + 1}};
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for (int[] d : direction) {
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int[][] copy = new int[visit.length][visit[0].length];
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for (int k = 0; k < visit.length; k++) {
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copy[k] = Arrays.copyOf(visit[k], visit[0].length);
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}
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if (dfs(board, word, d[0], d[1], index + 1, copy)) {
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return true;
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}
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}
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return false;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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src/main/java/leetcode/editor/cn/WordSearch.md
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src/main/java/leetcode/editor/cn/WordSearch.md
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<p>给定一个 <code>m x n</code> 二维字符网格 <code>board</code> 和一个字符串单词 <code>word</code> 。如果 <code>word</code> 存在于网格中，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p>
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<p>单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word2.jpg" style="width: 322px; height: 242px;" />
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<pre>
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<strong>输入：</strong>board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
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<strong>输出：</strong>true
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</pre>
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<p><strong>示例 2：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg" style="width: 322px; height: 242px;" />
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<pre>
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<strong>输入：</strong>board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
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<strong>输出：</strong>true
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</pre>
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<p><strong>示例 3：</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/word3.jpg" style="width: 322px; height: 242px;" />
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<pre>
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<strong>输入：</strong>board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
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<strong>输出：</strong>false
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>m == board.length</code></li>
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<li><code>n = board[i].length</code></li>
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<li><code>1 <= m, n <= 6</code></li>
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<li><code>1 <= word.length <= 15</code></li>
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<li><code>board</code> 和 <code>word</code> 仅由大小写英文字母组成</li>
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</ul>
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<p> </p>
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<p><strong>进阶：</strong>你可以使用搜索剪枝的技术来优化解决方案，使其在 <code>board</code> 更大的情况下可以更快解决问题？</p>
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<div><div>Related Topics</div><div><li>数组</li><li>回溯</li><li>矩阵</li></div></div>\n<div><li>👍 973</li><li>👎 0</li></div>
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