【Hot100】LeetCode—74. 搜索二维矩阵

• 原题链接： 74. 搜索二维矩阵

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1- 思路

二分

• ① 实现一个二分函数
• ② 对每行数组进行二分

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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;}
}

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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));}
}

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