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判断任意N*M数码有解的结论首先将矩阵存入n*m的一维数组，求去掉0以后的逆序数。 1.左右移动一次不改变逆序数奇偶性， 2.上下移动一次时: (1) 如果列数为奇数，逆序数奇偶性不变 (2) 如果列数为偶数，逆序数奇偶性改变一次，此时要统计始态和终态0的行数差的绝对值，若为偶数则始态和终态逆序数奇偶性相同，否则相反 Followed by:
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