[路飞]_二分查找以及算法题

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百度百科：在计算机科学中，二分搜索（英语：binary search），也称折半搜索（英语：half-interval search）、对数搜索（英语：logarithmic search），是一种在有序数组中查找某一特定元素的搜索算法。搜索过程从数组的中间元素开始，如果中间元素正好是要查找的元素，则搜索过程结束；如果某一特定元素大于或者小于中间元素，则在数组大于或小于中间元素的那一半中查找，而且跟开始一样从中间元素开始比较。如果在某一步骤数组为空，则代表找不到。这种搜索算法每一次比较都使搜索范围缩小一半。

可以说二分查找是一种非常高效的搜索算法

二分算法的基本结构

function binarySearch(arr, target) {
  var left = 0,
    right = ...
  while (...) {
    var mid = (left + right) >> 1
    if (arr[mid] === target) {
        ...
    }
    else if (target < arr[mid]) {
        ...
    }
    else if (target > arr[mid]) {
        ...
    }
  }
  return ...
}

基本的二分算法

寻找一个数

二分查找的裸题 704. 二分查找

function binarySearch (nums, target) {
    if (!nums.length) return -1
    var left = 0, right = nums.length - 1
    while (left <= right) {
        var mid = (left + right) >> 1
        if (nums[mid] === target) return mid
        else if (nums[mid] < target) left = mid + 1
        else if (nums[mid] > target) right = mid - 1
    }
    return -1
};
//nums [-1,0,3,5,9,12]
//target 9
binarySearch(nums, target) // 4

思考一个问题：如果数组中存在相同的元素比如[-1,0,3,3,5,9,12]，需要寻找3此时是希望返回左侧的还是右侧下标的元素既2或3

寻找左侧边界

function left_binarySearch (nums, target) {
    if (!nums.length) return -1
    var left = 0, right = nums.length - 1
    while (left < right) {
        var mid = (left + right) >> 1
        if (nums[mid] === target) right = mid
        else if (nums[mid] < target) left = mid + 1
        else if (nums[mid] > target) right = mid
    }
    return left
};
//nums [-1,0,3,3,5,9,12]
//target 3
left_binarySearch(nums, target) // 2

寻找右侧边界

function right_binarySearch (nums, target) {
    if (!nums.length) return -1
    var left = 0, right = nums.length - 1
    while (left < right) {
        var mid = (left + right) >> 1
        if (nums[mid] === target) left = mid + 1
        else if (nums[mid] < target) left = mid + 1
        else if (nums[mid] > target) right = mid
    }
    return left - 1
};
//nums [-1,0,3,3,5,9,12]
//target 3
right_binarySearch(nums, target) // 3

详细参考

小范围优化

function binarySearch (nums, target) {
    if (!nums.length) return -1
    var left = 0, right = nums.length - 1
    while (right - left > 3) { // 如果左右小于3
        var mid = (left + right) >> 1
        if (nums[mid] === target) return mid
        else if (nums[mid] < target) left = mid + 1
        else if (nums[mid] > target) right = mid - 1
    }
    // 小范围循环
    for (var i = left; i <= right; i++) {
        if (numbs[i] === target) return i
    }
    return -1
};
//nums [-1,0,3,5,9,12]
//target 9
binarySearch(nums, target) // 4

二分算法应用

633. 平方数之和

/**
 * @param {number} c
 * @return {boolean}
 */
var judgeSquareSum = function(c) {
    if(c === 0) return true
    for(var i = 0; i < Math.sqrt(c); i++) {
        var left = i, right = Math.floor(Math.sqrt(c))
        // 二分
        while(left <= right) {
            var mid = (left + right) >> 1
            var rst = i*i + mid*mid
            if(rst === c) return true
            else if(rst < c) left = mid + 1
            else if(rst > c) right = mid - 1
        }
    }
    return false
};

34. 在排序数组中查找元素的第一个和最后一个位置

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var searchRange = function (nums, target) {
  var ret = [-1, -1]
  ret[0] = binarySearch(nums, target)
  if (ret[0] == nums.length || nums[ret[0]] != target) {
    return [-1, -1]
  }
  ret[1] = binarySearch(nums, target + 1) - 1//查找第一个比target大的位置，前一位就是target结束位置。如果是最后一位返回 nums.length -1
  return ret
};
// 查找第一个大于等于x的值
var binarySearch = function (nums, x) {
  var head = 0,
    tail = nums.length - 1,
    mid
  while (tail - head > 3) {
    mid = (head + tail) >> 1
    if (nums[mid] >= x) tail = mid
    else head = mid + 1
  }
  for (var i = head; i <= tail; i++) {
    if (nums[i] >= x) return i
  }
  return nums.length //最大长度下一位方便求值
}

--未完待续--