LeetCode 74. Search a 2D Matrix

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.

分析

这是一道二分查找题, 只不过扩展到了二维. 需要注意循环判定条件以及对于lo和hi的修改. 这里我们对这两个变量起了不同的名字.

- lo hi 纵向 up down 横向 left right

外层循环寻找target可能位于哪一行, 而内层循环确定target是否存在.与我在这篇文章中介绍的题目类似, up为可能的行的最小下标, down为可能的行的最大下标加1. 如果target位于middle行两端数字构成的闭区间内, 则在此行内进行二分查找, 查找的结果即为函数返回值.

解答

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        int m = matrix.size();        if (m) {            int n = matrix[0].size();            if (n) {                int up = 0, down = m;                while (up < down) {                    int middle = up + (down-up) / 2;                    if (matrix[middle][0] <= target && target <= matrix[middle][n-1]) {                        int left = 0, right = n;                        while (left < right) {                            int mid = left + (right-left) / 2;                            if (matrix[middle][mid] == target) {                                return true;                            } else if (matrix[middle][mid] > target) {                                right = mid;                            } else {                                left = mid + 1;                            }                        }                        return false;                    } else if (matrix[middle][n-1] < target) {                        up = middle+1;                    } else if (matrix[middle][0] > target){                        down = middle;                    }                }                return false;            }            return false;        }        return false;    }};