74. Search a 2D Matrix

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本文介绍了一种高效的二维矩阵搜索算法，适用于每一行从左到右递增排列且每行首元素大于前一行尾元素的矩阵。通过二分查找法实现对目标值的有效定位。

Search a 2D Matrix

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.

解析

根据题意，如果把二维数组拉直，看成一列的话，很明显这是一个递增的数列。则很容易想到二分查找法。
同时，根据某个值所在的序号(index)，易计算其所在行和列。

行号=index/collength;
列号=index%collength;

代码

bool searchMatrix(int** matrix, int matrixRowSize, int matrixColSize, int target) {
    //二分查找
    int left=0;
    int right = matrixRowSize*matrixColSize-1;
    while(left<=right){
        int mid = (left+right)/2;
        int row = mid/matrixColSize;
        int col = mid%matrixColSize;
        if(matrix[row][col]==target){
            return true;
        }
        else if(matrix[row][col]>target){
            right=mid-1;
        }
        else{
            left=mid+1;
        }
    }
    return false;
}