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var a=[1,2,3,4,5,6];
' d. ~- j* ] J4 g% pvar b=[2,3,6]; G) A/ C; P+ I2 V( y2 T) `; _
var c =[];
; Y q5 g( I: h qfor(var i=0;i<a.length;i++){3 }: d' N# v, \/ [! c
var has = false;
9 ]. J. @( P! S* v, N! G, Bvar data = a[i];
) W n. `. l, n: w A//判断baic中是否有du元素data h. i, a3 H4 ^7 f! d. J
for(var j=0;j<c.length;j++){
' v7 A0 g, E2 E# \- s8 qif(data ==c[j]){. t# e8 P- v: h/ d* p+ C, k4 |
has =true;
Y+ z: J9 e" T+ t, u2 rbreak;
5 \9 z" z( O. k g& e# X' M" t: s6 T}) M( U& ]: m: Z
}5 I' Q" A& d3 N$ H$ V5 A
//如果zhi没有
# }8 U Z# _# U# B) ], ?if(has==false){5 I, z1 u" D$ j1 `3 u/ j
c.push(data);+ p: L0 h5 E/ m1 X! ^
}
1 A5 u+ c) V7 I% [- a}
" j/ T1 w+ E) t8 O8 U' ^8 Dfor(var i=0;i<b.length;i++){: g! s% Y3 x, ]- |- | H
bool has = false;1 \' D% {7 j- W
var data = a[i];
o: k& d2 h' m5 C9 o//判断c中是否有元素data( a) W0 G& _1 W: O% k2 s
for(var j=0;j<b.length;j++){
1 a4 ~2 O. A+ y- y/ u! O7 j2 {4 Oif(data ==b[j]){& ^2 V7 T+ I. z
has =true;' g8 q& s4 ?& e( h' u+ K
break;8 P5 S0 ]; w- S3 q. `
}" V4 @$ Y5 G) u, |$ K, @: ~0 [
}, f4 v: k0 g# k4 ?) @
//如果没有
2 t1 s: f& W. Bif(has==false){+ g3 Z& b/ e% V% i* {
c.push(data);
+ d( j) v5 L5 I1 M( I* V4 Z}, Y9 d4 d. d* \) d3 W6 A P( h
}
& v- k, U4 c( P2 x' r S% X" d. q//最后c就是结果 |
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