LeetCode Binary Search All In One

LeetCode Binary Search All In One

Binary Search

二分查找算法

https://leetcode-cn.com/problems/binary-search/

https://leetcode-cn.com/problems/binary-search/solution/er-fen-cha-zhao-by-leetcode/

复杂度分析

时间复杂度：\mathcal{O}(\log N)O(logN)。

空间复杂度：\mathcal{O}(1)O(1)。

LeetCode Binary Search Best Solutions in JavaScript

1. 位运算

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
const search = (nums, target) => {
  if(nums.length === 0){
    return -1;
  }
  let lo = 0;
  let hi = nums.length - 1;
  while(lo <= hi){
    // 位运算 12 >> 1 === 6
    const mid = lo + ((hi - lo) >> 1);
    if(nums[mid] === target){
      return mid;
    }else if(nums[mid] < target){
      lo = mid + 1;
    }else{
      hi = mid - 1;
    }
  }
  return -1;
};

1. 经典双指针

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    let left = 0;
    let right = nums.length-1
    let middle
    while(right >= left){
        middle = Math.floor((left+right)/2)
        const midval = nums[middle]
        if(midval === target) return middle;
        else if(midval > target) right = middle-1;
        else left = middle +1;
    }
    return -1
};

bad

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
  const findValue = (arr, target) => {
    let result = -1;
    let len = arr.length;
    let index = Math.floor(len / 2);
    let mid = arr[index];
    let leftArr = arr.slice(0, index)
    let rightArr = arr.slice(index + 1, len)
    if(mid === target) {
      result = nums.indexOf(mid);
    } else {
      if(mid > target) {
        // left
        result = findValue(leftArr, target)
      }
      if(mid < target) {
        // right
        result = findValue(rightArr, target)
      }
    }
    return result;
  }
  return findValue(nums, target);
};

refs

https://leetcode.com/problemset/all/

https://leetcode-cn.com/problemset/all/

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