LeetCode 78 Search a 2D Matrix

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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.
For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

Given target = 3, return true.

分析：

二分查找，关键在于怎么在二维数组上面进行一维二分查找。

长度为 m*n

mid元素是matrix[mid/n][mid%n]

这样就可以进行二分了。

public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        //检查参数        if(matrix==null || matrix.length==0 || matrix[0].length==0)            return false;        int m = matrix.length;        int n = matrix[0].length;        //将矩阵转化成一维数组进行二分        int start = 0;        int end = m*n-1;        while(start <= end){            int mid = (start+end)/2;            int val = matrix[mid/n][mid%n];//这行是关键            if(val == target)                return true;            else if(val < target)                start = mid+1;            else                end = mid-1;        }        return false;    }}

递归风格

public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix==null || matrix.length==0 || matrix[0].length==0)            return false;        int m = matrix.length;        int n= matrix[0].length;        return searchMatrix(matrix, target, 0, m*n-1);    }        public boolean searchMatrix(int[][] matrix, int target, int start, int end){        if(start > end)            return false;        int mid = (start+end)/2;        int x = mid/matrix[0].length;        int y = mid%matrix[0].length;//这两行是关键        if(matrix[x][y]==target)            return true;        if(matrix[x][y] < target)            return searchMatrix(matrix, target, mid+1, end);        else            return searchMatrix(matrix, target, start, mid-1);    }}

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