講解對(duì)象：C語言二分查找法

作者：融水公子 rsgz

前提:必須是有序的數(shù)據(jù)

left=0;    middle=(left+right);    right=9;

left>right  意味著找不到

#include<stdio.h>

#include<stdlib.h>

int main(int argc, const char* argv[]) {

int score[10], x, i, flag = 0;

int left = 0, middle, right = 9;

printf("input ten numbers please:\n");

for (i = 0; i<10; i++) {

scanf_s("%d", &score[i]);

}

printf("\n查找目標(biāo):\n");

scanf_s("%d", &x);

while (left <= right) {

middle = (left + right) / 2;

if (score[middle] == x) {

printf("數(shù)據(jù)存在  位置score[%d]\n", middle);

flag = 1;

break;

}

else if (score[middle]>x) {//數(shù)據(jù)在左邊 中間線賦值給右邊界

right = middle - 1;

}

else {                    //在右邊   中間線賦值給左邊界

left = middle + 1;

}

}

if (flag == 0) {

printf("數(shù)據(jù)不存在!\n");

}

system("pause");

return 0;

}