js binary search algorithm

js binary search algorithm

js 二分查找算法

二分查找, 前置条件

1. 存储在数组中
2. 有序排列

理想条件: 数组是递增排列，数组中的元素互不相同;

重排 & 去重

顺序: 递增排列/递减排列;
重复: 数组中存在相同的元素;

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2020-05-20
 * @modified
 *
 * @description 二分查找 binary-search
 * @augments
 * @example
 * @link
 *
 */

const log = console.log;

function binarySearch(data, key, preIndex = 0, debug = false){
  let len = data.length;
  // 向下取整
  let half = Math.floor(len/2);
   // 差值
  let diffValue = key - data[half];
  if(debug) {
    // preIndex 保留上次的索引值 indexOf
    log(`\npreIndex`, preIndex)
    log(`len`, len)
    log(`half`, half)
    log(`key`, key)
    log(`data[half]`, data[half])
    // number - undefined == NaN
    log(`diffValue`, diffValue)
  }
  if (diffValue > 0) {
    preIndex = preIndex + half + 1;
    let right = data.slice(half + 1);
    return binarySearch(right, key, preIndex);
  } else if (diffValue < 0) {
    let left = data.slice(0, half);
    return binarySearch(left, key, preIndex);
  }else if (diffValue === 0) {
    return preIndex + half;
  } else {
    return -1;
  }
}



let v4 = binarySearch([1,3,5,7,9], 9);
log(`v4`, v4)
// 4

let v11 = binarySearch([1,3,5,7,9], 11);
log(`v11`, v11)
// -1

// binarySearch(5, [1,2,3,4,5,6,7,8]);
// // 4
// binarySearch(5, [1,3,5,7,9]);
// // 2
// binarySearch(1, [1,3,5,7,9]);
// // 0
// binarySearch(7, [1,3,5,7,9]);
// // 3
// binarySearch(9, [1,3,5,7,9]);
// // -1

167. 两数之和 II - 输入有序数组

https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/

https://leetcode.cn/problems/two-sum-ii-input-array-is-sorted/

https://leetcode-solution-leetcode-pp.gitbook.io/leetcode-solution/easy/167.two-sum-ii-input-array-is-sorted

BST

function binarySearch(target,arr,start,end) {
    if( start > end){
        return -1
    }
    var start   = start || 0;
    var end     = end || arr.length-1;
    var mid = parseInt(start + (end-start)/2);
    if(target==arr[mid]){
        return mid;
    }else if(target>arr[mid]){
        return binarySearch(target, arr, mid+1, end);
    }else{
        return binarySearch(target, arr, start, mid-1);
    }
    return -1;
}

binarySearch(9,[1,2,3,4,5,7,8])
// -1

https://www.jianshu.com/p/eef65b21ace0

Big O

refs

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