Search a 2D Matrix & II

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.

 

• 在数组中，查找是否存在target这个数

bool searchMatrix(int** matrix, int matrixRowSize, int matrixColSize, int target) {
    int i = 0;
    for(; i < matrixRowSize; i++) 
    {
        if(matrix[i][matrixColSize - 1] >= target)
            break;
    }
    if(i == matrixRowSize)
        return 0;
    for(int j = 0; j < matrixColSize; j++)
    {
        if(matrix[i][j] == target)
            return 1;
    }
    return 0;
}

• 先查找每行最后一个值，对比target；得到哪行后对比该行

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

 

bool searchMatrix(int** matrix, int matrixRowSize, int matrixColSize, int target) {
    int i = 0, j = matrixColSize - 1;
    while(i < matrixRowSize && j >= 0)
    {
        if(matrix[i][j] == target)
            return 1;
        else if(matrix[i][j] > target)
            --j;
        else
            ++i;
    }
    return 0;
}

• i代表行，j代表列，选择第一行，最后一列，如果该值大于target则减少一列；否则增加一行，时间为O(m + n)
• 从左下或右上开始比较
• 可以每一行都用二分法查找，或者每一列进行二分查找，时间为m*lg(n),或者n*lg(m)但是当m，n变大是会大于上面的方法