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【2019-03-29】给定一个非空整数数组,除了某个元素只出现一次以外,其余每个元素均出现两次。找出那个只出现了一次的元素。 #12

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Pearyman opened this issue Mar 29, 2019 · 5 comments

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@Pearyman
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example:

in : [4,1,2,1,2]
output: 4

in : [1 , 1, 2]
output: 2

@Alvinte
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Alvinte commented Mar 29, 2019

arr.reduce((i,v)=>i^v,0)

@Pearyman
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for (let i=0;i<nums.length;i++){
    if(nums.indexOf(nums[i])===nums.lastIndexOf(nums[i])){
      return nums[i]
    }
  }

@xumingjie1658
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for (let i=0;i<nums.length;i++){
    if(nums.indexOf(nums[i])===nums.lastIndexOf(nums[i])){
      return nums[i]
    }
  }

for (let i=0;i<nums.length;i++){
if(i===nums.lastIndexOf(nums[i])){
return nums[i]
}
}

@Pearyman
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for (let i=0;i<nums.length;i++){
    if(nums.indexOf(nums[i])===nums.lastIndexOf(nums[i])){
      return nums[i]
    }
  }

for (let i=0;i<nums.length;i++){
if(i===nums.lastIndexOf(nums[i])){
return nums[i]
}
}

@xumingjie1658 不对吧 兄弟 你这验证了么

@xumingjie1658
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for (let i=0;i<nums.length;i++){
    if(nums.indexOf(nums[i])===nums.lastIndexOf(nums[i])){
      return nums[i]
    }
  }

for (let i=0;i<nums.length;i++){
if(i===nums.lastIndexOf(nums[i])){
return nums[i]
}
}

@xumingjie1658 不对吧 兄弟 你这验证了么

好像是不对...

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