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//顺序查找//时间复杂度:O(n)int SequenceSearch(int a[], int value, int n){ int i; for(i=0; ivalue) high = mid-1; //从前半部分查找 else low = mid+1; //从后半部分查找 } return -1;}//二分查找,递归版本int BinarySearch2(int a[], int value, int low, int high){ int mid = low+(high-low)/2; if(a[mid]==value) return mid; if(a[mid]>value) return BinarySearch2(a, value, low, mid-1); if(a[mid] value) high = mid-1; else low = mid+1; } return -1;}//斐波那契查找
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