Leetcode 112627：盛水最多容器-删除有序数组重复项-移除元素

Posted 2022-12-04 hello，是翠花呀

自己从0到1的思路，仅供参考：
一：盛最多水的容器:

/**
 * @param number[] height
 * @return number
 */
var maxArea = function(height) 
    var arr = [...height]
    arr.sort(function (a, b) 
      if (a - b > 0) 
        return 1
       else 
        return -1
      
    )
  
    var big = arr[arr.length - 1]
    var bigIndex = height.indexOf(big)
    var result = 0
    var leftH = 0
    var leftIndex = 0
    var rightH = height[height.length - 1]
    var rightIndex = height.length - 1
  
    for (var j = height.length - 1; j >= bigIndex; j--) 
        rightH = height[j]
        rightIndex = j

        for (var i = 0; i <= bigIndex; i++) 
            leftH = height[i]
            leftIndex = i
          if (rightH !== leftH) 
            var num = Math.abs((rightIndex - leftIndex) * ((rightH - leftH)>0 ? leftH : rightH))
           else 
            var num = Math.abs((rightIndex - leftIndex) * rightH)
          
          if (num > result) 
            result = num
          
        
    
    return result
;

二.删除有序数组中的重复项

/**
 * @param number[] nums
 * @return number
 */
var removeDuplicates = function(nums) 
  let i = 0
  let obj = 
  let resultLen = 0

  for (i; i < nums.length; i++) 
    if (!obj[nums[i]]) 
      obj[nums[i]] = 1
      nums[resultLen] = nums[i]
      resultLen++
    
  

  return resultLen
;

三.移除元素

/**
 * @param number[] nums
 * @param number val
 * @return number
 */
var removeElement = function(nums, val) 
  let len = nums.length
  let l = 0 // 因为只有不是被踢元素才unshift，所以不能用i来标识当前循环位置

  for (let i = 0; i < len; i++) 
    if (nums[i+l] !== val) 
      nums.unshift(nums[i+l])
      l++
     else 

    
  
  return l
;