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一个想法,o(n)的找到x和y的最大值,如果xmax和ymax出于同一个猴子,则候选数必为1,因为别的猴子的x或者y肯定会有至少一个比这个猴子小的; 如果没有在同一个猴子身上,则判断有多少个x==xmax||y==ymax) ; 实现了就是o(n)的,如有错误,请各位大牛指正…… Followed by:
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