冒泡basic程序
时间: 2024-02-27 17:55:22 浏览: 36
.getFontMetrics().descent;
canvas.drawText(text, x, y, mPaint);
}
}
if (lastHeaderPosition >= 0 && lastHeaderPosition < itemCount - 1) {
View firstChild = parent.getChildAt(0);
int firstVisible以下是一个基本的冒泡排序的BASIC程序,可以在BASIC编译器中运行:
```
10 DIMPosition = parent.getChildAdapterPosition(firstChild);
View lastHeader = parent.getChildAt(childCount - 1);
int lastHeaderTop = A(10)
20 INPUT "请输入10个数:", A(1), A(2), A(3), A(4), lastHeader.getTop();
int nextHeaderPosition = getNextHeaderPosition(lastHeaderPosition);
if (nextHeaderPosition == firstVisiblePosition A(5), A(6), A(7), A(8), A(9), A(10)
30 FOR I = ) {
int lastHeaderBottom = lastHeader.getBottom();
int overlayTop = Math.max(firstChild.getTop(), lastHeaderBottom1 TO 9
40 FOR J = 1 TO 10 - I
50 IF A(J) > A(J + 1));
int overlayBottom = Math.min(lastHeaderTop + lastHeader.getHeight(), parent.getBottom());
mTvOverlay.setText(mAppInfos.get(nextHeaderPosition).firstLetter);
showOverlay(overlayTop, overlayBottom);
} else {
hideOverlay();
}
THEN
60 T = A(J)
70 A(J) = A(J + 1)
80 A(J + 1) = T
} else {
hideOverlay();
}
}
private int getNextHeaderPosition(int position) {
String currentLetter = m90 END IF
100 NEXT J
110 NEXT I
120 PRINT "排序后的结果为:"
130 FOR I = 1 TOAppInfos.get(position).firstLetter;
for (int i = position + 1; i < mAppInfos.size(); i++) {
10
140 PRINT A(I)
150 NEXT I
```
程序通过输入10个数,使用冒泡排序算法对这 AppInfo appInfo = mAppInfos.get(i);
if (!TextUtils.equals(appInfo.firstLetter, currentLetter)) {
return i;
}
}
return mAppInfos.size();
}
@Override
public void getItemOffsets(@NonNull Rect out些数进行排序,最终输出排序后的结果。程序逐个比较相邻的数,如果前面的数比Rect, @NonNull View view, @NonNull RecyclerView parent, @NonNull RecyclerView.State state) {
int position = parent.getChildAdapterPosition(view后面的数大,则交换它们的位置。重复进行这样的比较和交换,直到所有的数都排好序为止。