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.
题目大意：给一个m*n矩阵，每一行都是递增有序，逐行也是递增的，要求设计一个高效算法检查目标元素是否存在于此矩阵中。
解题思路：可以把整个矩阵展开，就是一个长数组，数组长度为M*N，用二分查找即可，就是需要把二分查找的位置转化为矩阵下标。假设有row行，col列，那么key对应的矩阵中的元素应该是matrix[key/col][key%col]，这样就转为二分查找了。
Talk is cheap>>
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix[0][0] > target)
return false;
int rowLen = matrix.length;
int colLen = matrix[0].length;
int low = 0, high = rowLen * colLen - 1;
while (low <= high) {
int mid = (low + high) >>> 1;
int x = mid / colLen;
int y = mid % colLen;
if (matrix[x][y] > target) {
high = mid - 1;
} else if (matrix[x][y] < target) {
low = mid + 1;
} else {
return true;
}
}
return false;
}