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This article mainly explains "how to choose sorting and its optimization". The content of the explanation in the article is simple and clear, and it is easy to learn and understand. let's study and learn "how to choose sorting and its optimization".

Let's first take a look at the original version.

For (int I = 0; I

< beg.length - 1; i++) {for (int j = i + 1; j < beg.length; j++) {if (beg[i] >

= beg [j]) {int temp = beg [j]; beg [j] = beg [I]; beg [I] = temp;}} for (int k = 0; k

< beg.length; k++) { System.out.print(beg[k] + ",");} 我们可以发现外面的循环每次都是第i个数和剩下的length-i-1个数做比较 如下图

Optimization:

Each time we traverse to find the maximum and minimum values, then we will reduce the number of loops by 1 prime 2.

For (int I = 0; I

< (beg.length - 1) / 2; i++) {int min = i;int max = i;for (int j = i + 1; j < beg.length - i; j++) { min = beg[i] >

= beg [j]? J: min; max = beg [I] > = beg [j]? Max: J;} if (min + max = = beg.length-1) {int temp3 = beg [beg.length-1-I]; beg [beg.length-1-I] = beg [I]; beg [I] = temp3;} else {int temp = beg [min]; int temp2 = beg [max]; beg [min] = beg [I]; beg [max] = beg [beg.length-1-I]; beg [I] = temp;beg [beg.length-1-I] = temp2 }} for (int k = 0; k

< beg.length; k++) { System.out.print(beg[k] + ",");} 最外层遍历少了1/2 而里层遍历又少随着最大值和最小值的缩小而缩小区间

Here we need to consider the case of the extreme value, that is, the exchange of the maximum and minimum values. For example, 1 and 9 in the figure happen to be the maximum and minimum values.

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