Given an array of characters`chars`, compress it using the following algorithm:

Begin with an empty string s. For each group of**consecutive repeating characters** in`chars`:

- If the group's length is`1`, append the character to`s`.

- Otherwise, append the character followed by the group's length.

The compressed string`s`**should not be returned separately**, but instead, be stored**in the input character array**`chars`. Note that group lengths that are 10 or longer will be split into multiple characters in`chars`.

After you are done**modifying the input array**, return the new length of the array.

You must write an algorithm that uses only constant extra space.

**Example 1:**

**Example 2:**

String Compression

**Example 3:**

package Solution

funcSolution(xbool)bool {

funcSolution(chars []byte)int {

length:=len(chars)

iflength==0 {

}

nowByte,sameCount:=chars[0],1

foridx:=1;idx<=length;idx++ {

ifidx<length&&chars[idx]==nowByte {

}

chars[targetIndex]=nowByte

ifidx<length {

nowByte=chars[idx]

}

ifsameCount==1 {

}

for_,b:=rangefmt.Sprintf("%d",sameCount) {

chars[targetIndex]=byte(b)

}

}

}

funcTestSolution(t*testing.T) {

cases:= []struct {

inputs []byte

expect1 []byte

}{

{"TestCase",true,true},

{"TestCase",true,true},

{"TestCase",false,false},

{"TestCase1",[]byte{'a','a','b','b','c','c','c'}, []byte{'a','2','b','2','c','3'},6},

{"TestCase2",[]byte{'a'}, []byte{'a'},1},

{"TestCase3",[]byte{'a','b','b','b','b','b','b','b','b','b','b','b','b'}, []byte{'a','b','1','2'},4},

}

t.Fatalf("expected: %v, but got: %v, with inputs: %v",

c.expect,got,c.inputs)

}

if!reflect.DeepEqual(c.inputs[:got],c.expect1) {

t.Fatalf("expect %v but got: %v",c.expect1,c.inputs[:got])

}

})

}

}

funcBenchmarkSolution(b*testing.B) {

}

funcExampleSolution() {

}