• 原题链接： 74. 搜索二维矩阵

1- 思路

二分

• ① 实现一个二分函数
• ② 对每行数组进行二分

2- 实现

⭐74. 搜索二维矩阵——题解思路

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        for(int[] m:matrix){
            if(binarySearch(m,target)){
                return true;
            }
        }
        return false;
    }


    // 二分左区间
    public boolean binarySearch(int[] nums,int target){
        int left = 0;
        int right = nums.length-1;
        while(left<=right){
            int mid = (left+right)/2;
            if(nums[mid]<target){
                left = mid +1 ;
            }else if (nums[mid] > target){
                right = mid-1;
            }else{
                return true;
            }
        }
        return false;
    }
}

3- ACM 实现

public class binarySearch {


    public static boolean isHave(int[][] nums ,int target){
        for(int[] n:nums){
            if(binarySearch(n,target)){
                return true;
            }
        }
        return false;
    }

    public static boolean binarySearch(int[] nums,int target){
        int left = 0;
        int right = nums.length;
        while(left<=right){
            int mid = (left+right)/2;
            if(nums[mid]<target){
                left = mid+1;
            }else if(nums[mid] > target){
                right = mid-1;
            }else{
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String input = sc.nextLine();
        input = input.substring(2,input.length()-2);
        String[] parts = input.split("],\\[");
        String[] row = parts[0].split(",");

        int[][] nums = new int[parts.length][row.length];
        for(int i = 0 ; i < nums.length ;i++){
            String[] r = parts[i].split(",");
            for(int j = 0 ; j < r.length;j++){
                nums[i][j] = Integer.parseInt(r[j]);
            }
        }
        System.out.println("输入 target");
        int t = sc.nextInt();
        System.out.println("结果是"+isHave(nums,t));
    }
}