240. Search a 2D Matrix II

problem description

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.

algorithm thought

一个很有名的题目了，这是二分查找的二维形式。这里只要求每行每列的右边和下边都比他大，上面和左边都比他小。那么如果我们从左下角出发。每次都是往上或者右移动，这样就能保证要不就是往比他大的地方移动要不就是小的。

这也是二分查找的思路。要不就是往比mid大的值移动，要不就是小的值。

code

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

algorithm analysis

最多走到对角线，需要花费n+m步，时间复杂度是O(n+m)