题目:

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

思路:

先将数组排序,然后进行二分搜索。显然,中点的下标和中点的值相同时,说明从起始到中点没有错位,缺失数应该在数组后边。如果不相等,说明前面已经有错位,缺失数在左边。如果缺失数是最后一个的话,那整个数组都没有错位,则要返回最后一个加1。

/**
* @param {number[]} nums
* @return {number}
*/
var missingNumber = function(nums) {
if(nums.length==0){
return;
}
nums.sort(function(a,b){return a-b;});
var l=0,r=nums.length-1,middle=0;
while(l<r){
middle=l+Math.floor((r-l)/2);
if(middle!=nums[middle]){
r=middle-1;
}
if(middle==nums[middle]){
l=middle+1;
}
} return nums[l]==l?nums[l]+1:l
};

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