代码挑战：找到单一数字-Go
【摘要】 Given a non-empty array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without usi...
Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1]
Output: 1
Example 2:
Input: [4,1,2,1,2]
Output: 4方案1：
func singleNumber(nums []int) int {
result := nums[0]
for i := 1; i < len(nums); i++ {
result = result ^ nums[i]
}
return result
}
方案2：
func singleNumber(nums []int) int {
m := make(map[int]int)
for i := 0; i < len(nums); i++ {
if _, ok := m[nums[i]]; ok {
delete(m, nums[i])
} else {
m[nums[i]] = nums[i]
}
}
result := 0
for _, result := range m {
return result
}
return result
}
测试案例：
func Test_singleNumber(t *testing.T) {
type args struct {
nums []int
}
tests := []struct {
name string
args args
want int
}{
{"test1", args{nums: []int{2, 2, 1}}, 1},
{"test2", args{nums: []int{4, 1, 2, 2, 1}}, 4},
}
for _, tt := range tests {
t.Run(tt.name, func(t *testing.T) {
if got := singleNumber(tt.args.nums); got != tt.want {
t.Errorf("singleNumber() = %v, want %v", got, tt.want)
}
})
}
}
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