74. Search a 2D Matrix(js)

74. 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.

Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false
题意:给定一个s形排好序的二维数组,问我们数组中是否存在目标值target
代码如下:
/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
//比较数组行的第一项,若大于target,则遍历下一行,小于target,二分法查找本行
var searchMatrix = function(matrix, target) {
    
    
    if(matrix.length===0 || matrix===null || matrix[0].length===0) return false;
    var low=0;
    var high=matrix.length-1;
    
    while(low<=high){
        var mid=parseInt((low+high)/2);
        if(matrix[mid][0]===target) return true;
        else if(matrix[mid][0]>target) high=mid-1;
        else low=mid+1;
    }
    //找到可能存在target行
    var row=high;
    if(row<0) return false;
    
    low=0;
    high=matrix[0].length-1;
    while(low<=high){
        var mid=parseInt((low+high)/2);
        if(matrix[row][mid]===target) return true;
        else if(matrix[row][mid]>target) high=mid-1;
        else low=mid+1;
    }
    return false;


}
原文地址:https://www.cnblogs.com/xingguozhiming/p/10567877.html