LeetCode 74. Search a 2D Matrix

sunturing

于 2017-07-10 02:10:07 发布

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分类专栏：数据结构和算法

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本文介绍了一种高效的搜索算法，用于在一特殊性质的二维数组中查找指定值。该矩阵的每一行从左到右递增排序，并且每行的第一个元素大于前一行的最后一个元素。文章提供了一个具体的例子和实现代码。

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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.

在一个有序的二维数组里面查找一个值，思路是先对行二分查找，再对这行二分查找，不过我的代码只二分查找了行，遍历了这一行也能AC。。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length==0||matrix[0].length==0)return false;
        int row = matrix.length-1;
        int col = matrix[0].length-1;
        int l = 0;
        int r = row;
        int m = (l+r)/2;
        int c = 0;
        while(l<=r){
            if(matrix[m][0]>target&&m!=0&&matrix[m-1][0]<=target){
                c = m-1;
                break;
            }
            if(matrix[m][col]<target&&m!=row&&matrix[m+1][col]>=target){
                c = m+1;
                break;
            }
            if(matrix[m][0]>target){
                r = m-1;
                m = (l+r)/2;
            }
            else{
                l = m+1;
                m = (l+r)/2;
            }
        }
        for(int i=0;i<=col;i++){
            if(matrix[c][i]==target)return true;
        }
        return false;
    }
}