LeetCode：Search a 2D Matrix

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推荐参照：Leetcode题目难度等级及面试频率总结

题目描述：Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer of the previous row.

思路一：

首先，定位target值所在行，然后对该行进行二分查找即可。

    public boolean searchMatrix(int[][] matrix, int target) {        boolean isExist = false;        int row = matrix.length;//行数        int col = matrix[0].length;//列数        int target_row = 0;        int i = col - 1;        if(row * col == 0)            return false;        for(int j = 0 ;j < row; j ++){            ///寻找目标值所在行数            if(target >= matrix[j][0] && target <= matrix[j][i]){                isExist = true;                target_row = j;                break;            }        }        if(!isExist)            return false;        //在指定行进行二分查找        int low = 0;        int high = col - 1;        while(low <= high){            int mid = (low+high)/2;            if(matrix[target_row][mid]>target)                high = mid -1;            else if(matrix[target_row][mid] == target)                return true;            else                low = mid + 1;        }        return false;    }

思路二：

首先选取右上角数字，如果该数字等于要查找的数字，查找过程结束；如果该数字大于要查找的数字，去掉此数字所在列；如果该数字小于要查找的数字，则去掉该数字所在行。重复上述过程直到找到要查找的数字，或者查找范围为空。

import java.util.*;public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix == null || matrix.length == 0 || matrix[0].length == 0)            return false;        int i = 0, j = matrix[0].length - 1;        while(i < matrix.length && j >= 0){            if(matrix[i][j] == target)                return true;            else if(matrix[i][j] < target)                i++;            else                j--;        }        return false;    }}

思路三：

把二维的表格看成是一维的数组

import java.util.*;public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix == null || matrix.length == 0 || matrix[0].length == 0)            return false;        int row = matrix.length;        int col = matrix[0].length;        int i = 0, j = row * col - 1;        while(i <= j){            int mid = i + (j - i) / 2;            if(matrix[mid / col][mid % col] == target)                return true;            else if(matrix[mid / col][mid % col] < target)                i = mid + 1;            else                j = mid - 1;        }        return false;    }}

Any comments greatly appreciated.

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