Code & Func

第96天。

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:

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

Given target = 5, return true.

Given target = 20, return false.

以前好像看过这道题，但是应该嫌麻烦没做，今天做了一下，感觉好像也不是很难的样子，二分查找的升级版（在2维情况下）：

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bool searchMatrix(vector<vector<int>>& matrix, int target) {
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    int n = matrix.size();
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    if (n == 0) return false;
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    int m = matrix[0].size();
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    return searchMatrix(matrix,0,n-1,0,m-1,target);
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}
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bool searchMatrix(vector<vector<int> > &matrix, int xlow, int xhigh, int ylow, int yhigh, int target) {
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    //cout << xlow << " " << xhigh << endl
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    //   << ylow << " " << yhigh << endl;
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    if (xlow > xhigh || ylow > yhigh) return false;
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    int xmid = (xlow + xhigh)/2, ymid = (ylow + yhigh)/2;
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    if (matrix[xmid][ymid] == target) return true;
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    else if (matrix[xmid][ymid] < target)
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        return searchMatrix(matrix,xmid + 1, xhigh, ylow, yhigh,target) ||
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                searchMatrix(matrix,xlow, xhigh, ymid + 1, yhigh, target);
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    else
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        return searchMatrix(matrix, xlow, xmid-1, ylow, yhigh,target) ||
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                searchMatrix(matrix,xlow, xhigh, ylow, ymid-1, target);
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}

为什么dicuss的解法大多都是:

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bool searchMatrix(vector<vector<int>>& matrix, int target) {
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    int i = 0;
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    int j = matrix[0].size() - 1;
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    while(i < matrix.size() && j >= 0) {
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        if(matrix[i][j] == target)
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            return true;
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        if(matrix[i][j] < target)
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            i++;
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        else
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            j--;
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    }
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    return false;
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}