01.06 leetcode interview questions. String Compression (Python) (simple)

Title:

Solution:

1. After a string variable is provided to store the compressed temp1, calculation of a variable number of current occurrences of the letter count (initial value is 0);
2. Temp1 initialized to S [0], count plus 1; cycles followed by the string S; and if the current character is equal to the previous character, the current character repetitions + 1; if not equal, then the current character temp1 was added, the current character set number string;
3. Finally temp1 that is required of string compression.

code show as below:

class Solution:
    def compressString(self, S):

    	if S == '':
    		return S

    	#定义一个临时存储的空间
    	temp1 = ''

    	count = 0

    	if temp1 == '':
    			temp1 += S[0]
    			count += 1

    	for i in range(1,len(S)):


    		if S[i] != S[i - count]:			

    			
    			temp1 += str(count)
    			temp1 += S[i]


    			count = 1

    		else:

    			count += 1

    	temp1 += str(count)

    	if len(temp1) < len(S):
    		return temp1
    	else:
    		return S