【054-Spiral Matrix(螺旋矩阵)】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】原题
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. For example, Given the following matrix:
[ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ]]
You should return [1,2,3,6,9,8,7,4,5].
题目大意
给定一个m*n的矩阵,输入所有元素的螺旋顺序。
解题思路
使用计算输出的方法,先处理上面一行,,再处理右边一列,再处理下面一行,再处理左边一列,一直这样操作,直到所有的元素都处理完。
代码实现
算法实现类
import java.util.ArrayList;import java.util.List;public class Solution {public List<Integer> spiralOrder(int[][] matrix) {List<Integer> result = new ArrayList<>(50);if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {return result;}// 只有一行的情况if (matrix.length == 1) {for (int i : matrix[0]) {result.add(i);}return result;}// 只有一列的情况if (matrix[0].length == 1) {for (int i = 0; i < matrix.length; i++) {result.add(matrix[i][0]);}return result;}// 计算有多少圈int row = matrix.length;int col = matrix[0].length;int cycle = row < col ? row : col;cycle = (cycle + 1) / 2;int round = 0; // 记录当前是第几圈int left = 0;int right = matrix[0].length – 1;int top = 0;int down = matrix.length – 1;int total = col*row;int count = 0;while (round < cycle) {// 上面一行for (int i = left; i <= right && count < total; i++) {count++;result.add(matrix[round][i]);}top++; (int i = top; i <= down && count < total; i++) {count++;result.add(matrix[i][col – round – 1]);}right–;// 底下一行for (int i = right; i >= left && count < total; i–) {count++;result.add(matrix[row – round – 1][i]);}down–;// 左边一列for (int i = down; i >= top && count < total; i–) {count++;result.add(matrix[i][round]);}left++;round++;}return result;}}评测结果
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