请问在java数组中怎么判断奇数偶数的个数？ 在java怎样遍历数组中奇数的个数
java\u600e\u4e48\u5224\u65ad\u4e00\u4e2a\u6570\u662f\u5947\u6570\u8fd8\u662f\u5076\u6570\uff1f\u80fd\u591f\u88ab2\u6574\u9664\u7684\u6574\u6570\u662f\u5076\u6570\uff0c\u4e0d\u80fd\u88ab2\u6574\u9664\u7684\u662f\u5947\u6570\u3002
i % 2 == 1 ? "\u5947\u6570" : "\u5076\u6570";
\u4ee5\u4e0a\u7a0b\u5e8f\u5982\u679ci\u4e3a\u8d1f\u6570\u7ed3\u679c\u4e3a\u5076\u6570\u3002
Java\u53d6\u4f59(%)\u7b97\u6cd5\u4ee3\u7801\u5982\u4e0b\uff1a
// dividend\u88ab\u9664\u6570 divisor \u9664\u6570public static int remainder(int dividend, int divisor){ return dividend - dividend / divisor * divisor;}
\u6240\u4ee5\u5f53\u8f93\u5165 -1\u65f6 \u8fd0\u7b97\u7ed3\u679c\u662f-1\uff0c\u5f53\u7136\u4e0d\u7b49\u4e8e1\u4e86\uff0c\u6240\u4ee5\u5b83\u88ab\u5224\u5b9a\u4e3a\u5076\u6570\u4e86\u3002
\u56e0\u4e3a\u6539\u4e3a\u5224\u65ad\u662f\u5426\u662f\u5076\u6570\u5373\u53ef
i % 2 == 0 ? "\u5076\u6570" : "\u5947\u6570";
Java\u662f\u4e00\u79cd\u53ef\u4ee5\u64b0\u5199\u8de8\u5e73\u53f0\u5e94\u7528\u7a0b\u5e8f\u7684\u9762\u5411\u5bf9\u8c61\u7684\u7a0b\u5e8f\u8bbe\u8ba1\u8bed\u8a00\u3002Java \u6280\u672f\u5177\u6709\u5353\u8d8a\u7684\u901a\u7528\u6027\u3001\u9ad8\u6548\u6027\u3001\u5e73\u53f0\u79fb\u690d\u6027\u548c\u5b89\u5168\u6027\uff0c\u5e7f\u6cdb\u5e94\u7528\u4e8ePC\u3001\u6570\u636e\u4e2d\u5fc3\u3001\u6e38\u620f\u63a7\u5236\u53f0\u3001\u79d1\u5b66\u8d85\u7ea7\u8ba1\u7b97\u673a\u3001\u79fb\u52a8\u7535\u8bdd\u548c\u4e92\u8054\u7f51\uff0c\u540c\u65f6\u62e5\u6709\u5168\u7403\u6700\u5927\u7684\u5f00\u53d1\u8005\u4e13\u4e1a\u793e\u7fa4\u3002
public static void main(String[] args) { int num=0; int[] arr=new int[]{1,3,4,5,6,7,8,9,0,12,13}; for (int i = 0; i < arr.length; i++) { if (arr[i]%2!=0) { num++; } } System.out.println(num); }
int mun[]={1,2,3,4,5,6,7,8,9};
int count=0;
for(int i=0;i<mun.length;i++){
if(mun[i]!=0&&mun[i]%2==0)//0除外，能被2整除的数就是偶数
count++;
}
System.out.println("偶数有："+count+"个");
public static void main(String[] args){
int[] a = {1,2,3,4,5};
int jishu = 0;
int oushu = 0;
for(int i = 0;i < a.length; i++){
if(a[i]%2==0){
oushu ++;
}else{
jishu ++;
}
}
System.out.println("奇数个数为:" + jishu);
System.out.println("偶数个数为:" + oushu);
}可以求出数据中奇偶数个数
对于int x，判断 x%2的值：1 - x是奇数；0 - x是偶数
然后遍历一遍数组，计数就可以了。
绛旓細int mun[]={1,2,3,4,5,6,7,8,9};int count=0;for(int i=0;i<mun.length;i++){ if(mun[i]!=0&&mun[i]%2==0)//0闄ゅ锛岃兘琚2鏁撮櫎鐨勬暟灏辨槸鍋舵暟 count++; } System.out.println("鍋舵暟鏈夛細"+count+"涓");
绛旓細public static void main (String args[]) { int[] nums = {5, 2, 45, 11, 13, 32, 7, 24, 63, 48};int odd=0, even=0;for (int i=0; i<nums.length; i++) { if (nums[i] % 2 == 0)even ++;else odd ++;} System.out.println("even:\t" + even);System.out....
绛旓細} System.out.println("闅忔満6浣嶆暟鏁扮粍锛"+Arrays.toString(arr)+"\n鍏朵腑锛屽鏁颁釜鏁帮細"+odd +"\n鍋舵暟涓暟锛" + even);} }
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