一天一道LeetCode#74. Search a 2D Matrix

Posted 2020-07-19 ZeeCoder

一天一道LeetCode

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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;
    }
};