3 回答
TA贡献1831条经验 获得超4个赞
我认为下面的实现将正确地完成工作:
int firstOccurrence(int[] sequence, int x) {
int min = 0;
int max = sequence.length - 1;
int result = -1;
while (min <= max)
{
// find the mid value and compare it with x
int mid = min + ((max - min) / 2);
// if x is found, update result and search towards left
if (x == sequence[mid]) {
result = mid;
max = mid - 1;
} else if (x < sequence[mid]) {
// discard right half
max = mid - 1;
} else {
// discard left half
min = mid + 1;
}
}
// return the leftmost index equal to x or -1 if not found
return result;
}
编辑:
更改计算 mid 的方式以避免较大的和溢出
// Previously, can overflow since we add two integer
int mid = (min + max) / 2;
// Now
int mid = min + ((max - min) / 2);
// Another way using the unsigned right shift operator
int mid = (low + high) >>> 1;
// The left operands value (low + high) is moved right
// by the number of bits specified (2 in this case) by the right operand and
// shifted values are filled up with zeros.
// The >>> treats the value as unsigned
TA贡献1809条经验 获得超8个赞
这是等同于从 C++ 进行的搜索lower_bound
。它返回小于您要查找的值的元素数。那将是第一次出现的索引,或者如果没有出现将插入的位置:
int numSmaller(int[] seq, int valueToFind)
{
int pos=0;
int limit=seq.length;
while(pos<limit)
{
int testpos = pos+((limit-pos)>>1);
if (seq[testpos]<valueToFind)
pos=testpos+1;
else
limit=testpos;
}
return pos;
}
请注意,每次迭代我们只需要进行一次比较。
链接的答案强调了以这种方式编写二进制搜索的几个优点。
TA贡献1744条经验 获得超4个赞
它认为它会帮助你
public static boolean binarysearch(int[] data, int target, int low, int high){
if(low>high){
System.out.println("Target not found");
return false;}
else{
int mid=(low+high)/2;
if(target==data[mid])
return true;
else if(target<data[mid])
return binarysearch(data, target, low, high);
else
return binarysearch(data, target, low, high);
}
}
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