一天一道LeetCode#74. Search a 2D Matrix

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(一)题目

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.

For example,

Consider the following matrix:

[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]

Given target = 3, return true.

(二)解题

剑指offer上的老题了,矩阵是排好序的,那么我们可以从其中找到规律。
从右上角(0,n)开始扫描,如果target比它大就往下找,如果小就往左边找。
具体看代码:

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size()==0) return false;
        int i = matrix.size()-1;
        int j = 0;
        while(i>=0 && j< matrix[0].size())
        {
            if(target==matrix[i][j]) return true;//找到
            if(target>matrix[i][j]) j++;//如果target大,就往下找
            else i--;//反之则往左找
        }
        return false;
    }
};

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LeetCode 74:Search a 2D Matrix

Leetcode 74: Search a 2D Matrix