LeetCode Num74_Search a 2D Matrix

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　　问题描述：　　

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.

　　释义：在给定的 m X n 二维数组中搜索，判断是否存在target，数组每行有序，非递减顺序，下一行的第一个元素比上一个元素的最后一个元素大，算法没有要求。

　　问题分析：

　　1. 可以把二维数组的看作是一个有序的一位数组，即按行追加到上一行的末尾位置处，开辟一个大小为m+n大小的以为数组，利用二分法检索target，算法时间复杂度为O(logn)，空间复杂度为O(m+n)。

2. 先顺序检索到target所在的行位置（row index），然后在这一行中利用二分查找进行检索，时间复杂度为O(n)，由于无需开辟额外数组，所以空间复杂度为O(1)；

3. 先二分检索到target所在的行位置（row index），然后在这一行利用二分检索，时间复杂度为O(log n)，空间复杂度为O(1)；

　　4. 确定边界条件和异常情况；

　　实现：

　　i.顺序检索行：

/*
* @brief: search the "target" in the "sorted" Matrix which
* the first integer of each row is greater than the last integer of the previous row.
* @authour: zhangjinxing jinxingbay@163.com
* @time: 2015 01 04, 21 15
* @edition: 1.0
* @time complexity: O(n), can be "{optimized}" to O(log n) by using "{Binary Search}"
*/
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
const unsigned int m = matrix.size();
if (0 == m) {// matrix is empty, row number == 0
return false;
}
const unsigned int n = matrix[0].size();
if (0 == n) {// matrix is empty, row number != 0, but column number == 0
return false;
}
if (target < matrix[0][0] || target > matrix[m-1][n-1]) {// target is not in the matrix
return false;
}

unsigned int i=0, j=0, irow=0;// irow is the index of the choosen row
for (i=0; i<m; ++i) {// find the "target" row number
irow = i;// change the irow in case of "target" is existed in the last row
if (matrix[i][0] == target) {// return true if find the "target" in the matrix
return true;
}
if (matrix[i][0] > target) {// increasing index "i" while matrix[i][0] larger than target, return the previous row "i-1"
irow = i - 1;
break;
}
}

if (target < matrix[irow][0] || target > matrix[irow][n-1]) {// target is not in the matrix row where row index == irow
return false;
}

for (j=0; j<n; ++j) {// find the "target" column number
if (matrix[irow][j] == target) {// return true if find the "target" in the matrix row index == irow
return true;
}// else, increasing index "i" while not equals to "target"
}
return false;// if can not find the "target" in the matrix, return false
}
};

分析
　　先进行边界条件判断会提高程序运行效率，提高可维护性；如编码加粗部分。

ii. 二分检索行
/*
* @brief: search the "target" in the "sorted" Matrix which
* the first integer of each row is greater than the last integer of the previous row.
* @authour: zhangjinxing jinxingbay@163.com
* @time: 2015 01 04, 21 35
* @edition: 2.0
* @time complexity: O(long n) by using "{Binary Search}"
*/
class Solution {
public:
bool searchMatrix(vector<vector<int> > &matrix, int target) {
const unsigned int m = matrix.size();
if (0 == m) {// matrix is empty, row number == 0
return false;
}
const unsigned int n = matrix[0].size();
if (0 == n) {// matrix is empty, row number != 0, but column number == 0
return false;
}
if (target < matrix[0][0] || target > matrix[m-1][n-1]) {// target is not in the matrix
return false;
}

unsigned int row=0;// irow is the index of the choosen row
unsigned int rbeg=0, rend=m-1, rmd=m/2;
while (rbeg<=rend) {
if (rbeg == rend) {
irow = rend;
break;
}
rmd = rbeg + (rend-rbeg + 1) / 2; // pay attention on the index computing when "begin index" and "end index" changing
// when using "Binary Search"
if (matrix[rmd][0] == target) {
return true;
}
if (matrix[rmd][0] > target) {// row index must be in the range "[rbeg, rmd-1]"
rend = rmd - 1;
continue;
}
if (matrix[rmd][0] < target) {
rbeg = rmd;// row index must be in the range "[rmd, rend]", attention "rmd" not "rmd+1"
continue;
}
}

if (target < matrix[irow][0] || target > matrix[irow][n-1]) {// target is not in the matrix row where row index == irow
return false;
}

unsigned int cbeg=0, cend=n-1, cmd=n/2;
while (cbeg<=cend) {
if (cbeg == cend) {
if (matrix[irow][cbeg] == target) {
return true;
} else {// can not find the "target" in the matrix,
// "target > matrxi[irow][0] && target < matrix[irow][n-1], but no one equal to target"
return false;
}
}
cmd = cbeg + (cend-cbeg+1) / 2; // pay attention on the index computing when "begin index" and "end index" changing
// when using "Binary Search"
if (matrix[irow][cmd] == target) {
return true;
}
if (matrix[irow][cmd] > target) {// "[cbeg, cmd-1]"
cend = cmd - 1;
continue;
}
if (matrix[irow][cmd] < target) {// "[{cmd+1}, cend]" not "cmd"
cbeg = cmd + 1;
continue;
}
}
return false;// if can not find the "target" in the matrix, return false
}
};
分析：
二分法使用时一定要注意分开的两个区间的下标。
总结
i. consider special cases completely, using the whole given conditions;
ii. build the bug-free coding fragment is difficult, to be there, I must do exercises, coding & thinking;
iii. pay attention to the index when using Binary Search method;
iv. special cases comes first and just "code" it.

-----One "Acceptance" a day, keep my mind awake.-----noted by bayingbf-----2015.01.05-----

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