"Algorithm" to find a two-dimensional array

Find a two-dimensional array of 0001

Title Description

(The length of each one-dimensional array of same), each row from left to right in order of ascending sort, to sort each column in a two-dimensional array in order of increasing from top to bottom. A complete function, enter such a two-dimensional array and an integer, it is determined whether the array contains the integer.

Topics address

• https://www.nowcoder.com/practice/abc3fe2ce8e146608e868a70efebf62e

Report problem-solving

Violence Act

This problem solution by the micro-channel public number 小猿刷题provided wrong, please correct me.

One by one through the array, if found return true

/**
 *  微信公众号"小猿刷题"
 */
public class Solution {
    public boolean Find(int target, int [][] array) {
        for(int i = 0; i < array.length; i ++){
            for(int j = 0; j < array[i].length; j++){
                if(target == array[i][j]){
                    return true;
                }
            }
        }
        return false;
    }
}

Local binary search

This problem solution by the micro-channel public number 小猿刷题provided wrong, please correct me.

Progressive traversal, binary search within each row.

/**
 *  微信公众号"小猿刷题"
 */
public class Solution {
    public boolean Find(int target, int [][] array) {
        for(int i = 0; i< array.length; i++){
            int low = 0;
            int high = array[i].length - 1;
            while (low <= high){
                int mid = (low + high) / 2;
                if(target > array[i][mid]){
                    low = mid + 1;
                } else if(target < array[i][mid]){
                    high = mid - 1;
                } else {
                    return true;
                }
            }
        }
        return false;
    }
}

Overall binary search

This problem solution by the micro-channel public number 小猿刷题provided wrong, please correct me.

A two-dimensional array from top to bottom, from left to right in increasing regularity, then select the upper or lower left corner of the element array[row][col]and target are compared.

• When targetless than the element array[row][col]when it targetmust be an element in arraythe left side of the row, namely col--;
• When targetelements greater than array[row][col]the time, then targetcertainly in the element arraybelow where the columns, namely row++;

/**
 *  微信公众号"小猿刷题"
 */
public class Solution {
    public boolean Find(int target, int [][] array) {
        int row = 0;
        int col = array[0].length - 1;
        while (row <= array.length -1 && col >= 0) {
            if(target > array[row][col]) {
                row ++;
            } else if(target < array[row][col]) {
                col --;
            } else {
                return true;
            }
        }
        return false;
    }
}