标签：dma 返回 整数 instead max val 替换 nbsp 存在
Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).
给定一个非空数组的整数，返回该数组中的第三个最大数。 如果不存在，返回最大数量。 时间复杂度必须在O（n）中。
Example 1:
Input: [3, 2, 1] Output: 1 Explanation: The third maximum is 1.
Example 2:
Input: [1, 2] Output: 2 Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: [2, 2, 3, 1] Output: 1 Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum.
思路其实很简单，设置三个变量，first second third 分别存储前三大数，遍历数组，找到前3大数，根据不同情况将nums[i]和first second third 进行替换，
最后返回最大值或者第三大数；
var thirdMax = function(nums) { var len = nums.length; var first,second,third; if(len < 3){ third = Math.max.apply(Math,nums); }else{ first = nums[0]; for(var i = 1;i < len;i++){ if(second === undefined){ if(nums[i] < first){ second = nums[i]; }else if(nums[i] > first){ second = first; first = nums[i]; } }else if(second !== undefined && third === undefined){ if(nums[i] < second){ third = nums[i]; }else if(nums[i] > first){ third = second; second = first; first = nums[i]; }else if(nums[i] > second && nums[i] < first){ third = second; second = nums[i]; } }else if(second !== undefined && third !== undefined){ if(nums[i] > third && nums[i] < second){ third = nums[i]; }else if(nums[i] > second && nums[i] < first){ third = second; second = nums[i]; }else if(nums[i] > first){ third = second; second = first; first = nums[i]; } } } } if(third === undefined){ return first; }else { return third; } };
标签：dma 返回 整数 instead max val 替换 nbsp 存在
原文地址：http://www.cnblogs.com/deerfig/p/6680388.html
