[LeetCode#240] Search a 2D Matrix II

This problem is still very easy, but it reminds me to repick up some skills I have abandoned for a long time.
Instant idea:
Since each column and row is sorted, we can take advantage of one of them. Go through each row one by one, for each row, we perform binary search. 
Note: pitfall in binary search.
1. wrong stop condition.
while (left < right) 
Input:
[[-5]], -5
Output:
false
Expected:
true

start: left = 0, right = 0
not able to enter the loop, not able to reach
int mid = (left + right) / 2;
if (nums[mid] == target)
    return true;
Fix: while (left <= right)

2. wrong policy in update left and right pointer.
if (nums[mid] == target)
    return true;
else if(nums[mid] > target) {
    right = mid;
} else {
    left = mid;
}

Input:
[[-1,3]], 1
start: left = 0, right = 1, 
enter loop: mid = 0;
nums[mid] < target
    left = mid = 0.
No updates at all, would lead to infinite loop.
Binary search is very important, you must remember the coding structure clearly.
Time Complexity: O(nlogn)

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null)
            throw new IllegalArgumentException("the reference of the matrix is null");
        if (matrix.length == 0 || matrix[0].length == 0)
            return false;
        for (int i = 0; i < matrix.length; i++) {
            if (binarySearch(matrix[i], target))
                return true;
        }
        return false;
    }
    
    
    private boolean binarySearch(int[] nums, int target) {
        if (nums[0] > target || target > nums[nums.length-1])
            return false;
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target)
                return true;
            else if(nums[mid] > target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }
        return false;
    }
}

Even though we have take advantage of sorted property of each row, how about each column?
Can we also take advantage of it? so as to reduce time complexity. 

Sure!!!
Great idea:
Start from up-right-most corner, then use the two sorted properties to exclude elements. 
principle:
1. iff matrix[x][y] > target, 
since y is sorted, target must not able to appear on y column, we should skip it. 
while ( y >= 1 && target < matrix[x][y] ) {
    y--;
}   

2. iff matrix[x][y] < target.
since matrix[x][y-1] must smaller than it, and matrix[x][y+1] is impossilbe. (we have done in the previous step)
the only possible direction is below. 

Use this invariant, if the target exist in the matrix, we must be able to find it.
for (int x = 0; x < matrix.length; x++) {
    while ( y >= 1 && target < matrix[x][y] ) {
        y--;
    }   
    if (matrix[x][y] == target)
    return true;
}

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null)
            throw new IllegalArgumentException("the reference of the matrix is null");
        if (matrix.length == 0 || matrix[0].length == 0)
            return false;
        int y = matrix[0].length-1;
        for (int x = 0; x < matrix.length; x++) {
            while ( y >= 1 && target < matrix[x][y] ) {
                y--;
            }   
            if (matrix[x][y] == target)
                return true;
        }
        return false;
    }
}