本人参考网上对冒泡算法的优化,再一次进行的小小优化,欢迎各位指点5 T, p& D* v5 d$ I7 U+ U$ X$ N
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- void bubble4(uint16_t *arr, uint16_t length) " o4 |% w# K0 O& }
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- uint16_t borden_right = length -1; //右边界初始值
8 D. b& s8 ~, A8 J' y, ? K - uint16_t borden_left = 0; //左边界初始值为
) u2 e$ }9 }9 J1 H, q% @ - uint16_t lastPos = 0; //记录右边界的值
' m/ n( j/ b4 }1 J, e5 o - uint16_t prePos = 0; //记录左边界的值8 }- C B! E# f* ]+ E
- uint16_t temp; //交换的中间变量9 K. d4 R; l6 h/ Z% X
- uint8_t flag, i, j; //是否有交换的标志 0 T( F, o5 J( Q" m; n
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- for(i = 0; i < length-1; i++) ( U$ _- d. B0 P/ a, s6 Z# z! N
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4 a" `: B5 Z2 ~) |6 I4 P - flag = 1;7 @% b7 S" E g, T
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- //正向找出最大值% }3 C2 T% ?$ I; N& L; U
- for( j = borden_left; j < borden_right; j++)3 u6 c# H' c0 r8 n+ q
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( R6 N! A0 x4 |) _ - if(arr[j] > arr[j+1]) w2 Y# h; S! H7 I7 g- q* R
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- temp = arr[j];
6 _- g$ k# r! W. y: l8 \# { - arr[j] = arr[j+1];
% U' d! i% j' o/ V' | - arr[j+1] = temp;: P+ e- X: Z4 ?4 {1 ]
- flag = 0;
- f4 N e9 N: W. s - lastPos = j;
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+ K4 Y# n, z$ F/ K - //正向没发生交换就证明数组已经排好了! I' l+ z7 c7 d) L6 \
- if(flag), ]! O" J, W+ \5 c7 L
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! _* o! ^* d$ j8 X$ B+ U& A - break;) e* W5 E* d8 U1 C5 C ?6 R/ Y
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9 \; J. n, `! i6 R0 G - borden_right = lastPos;2 X1 h1 W5 u9 x+ n# s
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& f- e/ H% {* f - //逆向找出最小值
' E: F2 k, h3 O$ X - for( j = borden_right; j > borden_left; j--)
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" B: p* Y4 h& e; e, E& }- e6 K& i# m - if(arr[j] < arr[j-1])
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& P5 e) \, S+ A' M8 Y l - temp = arr[j];
& d4 o' `7 L8 E+ w: o - arr[j] = arr[j-1];
; o9 W6 f' e1 m, Z; Z- s - arr[j-1] = temp;6 b% x; Z" G( R u/ L" K) z( b+ R
- flag = 0;; a. Z; ~/ \6 S/ n# E: S6 k
- prePos = j;- z: e( j- ]/ i* }( u
- }
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A) `6 z5 D- P; T4 P2 k- if(flag) 4 Y! c3 f) I- v8 b) y) [. _: F$ h
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, F6 H6 t K* z+ {$ H R - break;6 c6 [ W! B0 I
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- borden_left = prePos;
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- //每排序一次都把数组打印出来
% s6 u- d( D$ I. v - // for(j = 0; j < length; j++)4 c. Q9 c1 c* I6 ?: K
- // {$ i: S( k$ d$ s8 _6 q' [5 D4 \
- // USART_SendData(USART1, arr[j]);
# B4 C5 k3 p; Q( E" R/ K$ v5 J# \ - // while(USART_GetFlagStatus(USART1, USART_FLAG_TC) == RESET);
" G& s- E( K1 }2 s+ G1 Q4 ?, l2 s0 } - // }
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复制代码
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如果楼主能提供这些数据,我觉得更好