[C++]LeetCode: 61 Search a 2D Matrix
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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
.
思路:根据题意,矩阵每行都有序排列,并且逐行增加,所以我们可以把MN矩阵展开成一维有序数组。
num表示元素的序列号,从0开始。i, j表示矩阵中的行列坐标,从0开始。
Num = n*i + j; => i = Num /n ; j = Num % n; n表示矩阵的列。(画图可以得到)
接下来利用二分查找,找到target.
Attention: 注意将矩阵中的序列号和矩阵行列号的转换。
int mid = lo + (hi - lo) /2; int i = mid / n; int j = mid % n;
复杂度:O(log(N))
AC Code: (MY_CODE ONE_PASS)
class Solution {public: bool searchMatrix(vector<vector<int> > &matrix, int target) { //其实类似于将mn矩阵展开成一个数组然后查找,不过需要根据第几个元素,计算行列 //num表示元素的序列号,从0开始。i,j表示矩阵中的行列坐标,从0开始。 //Num = n*i + j; => i = Num /n ; j = Num % n; n表示矩阵的列。 bool ret = false; if(matrix.size() == 0 || matrix[0].size() == 0) return ret; int m = matrix.size(); int n = matrix[0].size(); int lo = 0; int hi = (m - 1) * n + (n - 1); // hi = m * n -1; while(lo <= hi) { int mid = lo + (hi - lo) /2; int i = mid / n; int j = mid % n; if(target > matrix[i][j]) { lo = mid + 1; } else if(target < matrix[i][j]) { hi = mid - 1; } else { ret = true; break; } } return ret; }};
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