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Algorithms/AlgorithmsTest/src/Num1_1_04/Num_1_04_11.java #8

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lutaoact opened this issue Jun 1, 2017 · 0 comments

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lutaoact commented Jun 1, 2017

howMany()方法，在最坏情况下，复杂度为O(n)，如果一个数组的所有元素都是指定的可以，那循环就会执行n次。
我觉得正确的做法应该是，参考前一题，先找出最小索引，再找出最大索引，这两个操作都是对数级的复杂度，然后直接就能求出结果。

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