Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.
Here are few examples.
[1,3,5,6]
, 5 → 2
[1,3,5,6]
, 2 → 1
[1,3,5,6]
, 7 → 4
[1,3,5,6]
, 0 → 0
我们使用二分查找来解决
// Given a sorted array and a target value,
// return the index where it would be if it
// were inserted in order.
// You may assume no duplicates in the array.
// Here are few examples.
/* linear search */
function findPosition1 (value, arr) {
for(var i=0; i<arr.length; ++i) {
if(arr[i] > value) {
return i;
}
}
return i;
}
function test() {
// 0 1 2 3
var arr = [1, 3, 5, 6];
console.log(findPosition1(10, arr));
}
test();
/* 二分查找 */
function binarySearch(value, arr) {
var len = arr.length,
mid = Math.floor(len / 2);
if(value == arr[mid]) {
return mid;
} else if(value > arr[mid]) {
return binarySearchAux(mid+1, len);
} else {
return binarySearchAux(0, mid);
}
// binary search auxiliary
function binarySearchAux(start, stop) {
/*
@start: the start of arr (included)
@stop: the stop of arr(not incluedd)
*/
if(start == stop) {
return start;
}
var mid = (start + stop) / 2;
if(value == arr[mid]) {
return mid;
} else if(value > arr[mid]) {
return binarySearchAux(mid+1, stop);
} else {
return binarySearchAux(start, mid);
}
}
}
function anotherTest() {
// 0 1 2 3 4
var arr = [1, 2, 8, 9, 10];
console.log(binarySearch(-10, arr));
}
anotherTest();
另外提供一个更加简洁的非递归版本
/* Here I will offer you a non-recursive version */
function binarySearch1(value, arr) {
var len = arr.length,
low = 0,
high = len,
mid = Math.floor((low + high) / 2);
while(low < high) {
if(arr[mid] == value) {
return mid;
} else if (arr[mid] < value) {
low = mid + 1;
} else {
high = mid;
}
mid = Math.floor((low + high) / 2);
}
return low;
}
function otherTest() {
var arr = [10, 20, 30, 40, 50];
console.log(binarySearch1(500, arr));
}
otherTest();