1.递归实现
function binarySearch(data, dest, start, end){
var end = end || data.length - 1,
start = start || 0,
m = Math.floor((start + end) / 2);
if(data[m] == dest){
return m;
}
if(dest < data[m]){
return binarySearch(data, dest, 0, m-1);
}else{
return binarySearch(data, dest, m+1, end);
}
return false;
}
var arr = [-34, 1, 3, 4, 5, 8, 34, 45, 65, 87];
binarySearch(arr,4);
//3
2.非递归实现
function binarySearch(data, dest){
var h = data.length - 1,
l = 0;
while(l <= h){
var m = Math.floor((h + l) / 2);
if(data[m] == dest){
return m;
}
if(dest > data[m]){
l = m + 1;
}else{
h = m - 1;
}
}
return false;
}
var arr = [-34, 1, 3, 4, 5, 8, 34, 45, 65, 87];
binarySearch(arr,4);
//3
以上是一个等比数列，n / 2^k = 1时，k就是查找的次数。即k=log2n，所以时间复杂度为O(logn)，这是一种非常高效率的算法。
3.数组中存在重复的数据，怎么找出元素最后一次出现的位置❓
和上边介绍的二分查找思路一样：
function binary_search(arr, key) {
var low = 0, high = arr.length - 1;
while (low <= high) {
var mid = low + ((high - low) >> 1);
if (arr[mid] > key) {
high = mid - 1;
} else if (arr[mid] < key) {
low = mid + 1;
} else {
if ((mid == arr.length - 1) || (arr[mid + 1] != key)) return mid;
else low = mid + 1;
}
}
return -1;
}
var arr = [5,13,19,21,21,37,56,64,75,80,88,92];
var result = binary_search(arr, 21);
console.log(result);
// 左开右闭式
function binarySearch(array, start, end, target) {
while (start < end) {
const mid = Math.floor(start + (end - start) / 2)
const middle = array[mid]
if (target > middle) {
start = mid + 1
} else {
end = mid
}
}
return start
}
console.log(binarySearch([1,2,22], 0, 3, 22))
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