【刷题-LeetCode】240. Search a 2D Matrix II

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.

解 从右上角到左下角，往左搜索变小，往下搜索变大，假设在位置 i, j处的元素值为n，搜索过程为：

• target == n ：找到，结束
• target > n：往下走，增加n
• target < n：往左走，降低n

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int m = matrix.size();
        if(m == 0)return false;
        int n = matrix[0].size();
        int i = 0, j = n-1;
        while(i < m && j >= 0){
            if(matrix[i][j] == target)return true;
            if(matrix[i][j] > target){
                j--;
            }else{
                i++;
            }
        }
        return false;
    }
};