240. Search a 2D Matrix II

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

class Solution {
public:
    //从右上节点开始查找：
    //1、如果右上点matrix[i][col] < target ,那么第i行都会小于target,直接丢掉第i行。
    //2、如果右上点matrix[i][col] > target,那么第col列都会被丢掉，因为整个第col列都会大于target。
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int row = matrix.size();
        if(row == 0) return false;
        int col = matrix[0].size();
        if(col == 0) return false;
        int i = 0,j=col-1;
        while(i<row && j >=0){
            if(matrix[i][j] == target) return true;
            else if(matrix[i][j] < target) i++;
            else j--;
        }
        return false;
    }
};

原文地址：https://www.cnblogs.com/wsw-seu/p/14018495.html