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
class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0) return false;
        int m = matrix[0].length;
        int n = matrix.length;
        int left = 0;
        int right = m * n - 1;
        while(left <= right){
            int mid = left + (right - left)/2;
            int value = matrix[mid/m][mid%m];
            if(target == value) return true;
            else if(target < value) right = mid - 1;
            else left = mid + 1;
        }
        return false;
    }
}

把matrix想象成sorted list,直接用二分法搜索即可,对应的坐标是关键

转载于:https://www.cnblogs.com/wentiliangkaihua/p/11511268.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值