848. Shifting Letters

We have a string S of lowercase letters, and an integer array shifts.

Call the shift of a letter, the next letter in the alphabet, (wrapping around so that 'z' becomes 'a'). 

For example, shift('a') = 'b', shift('t') = 'u', and shift('z') = 'a'.

Now for each shifts[i] = x, we want to shift the first i+1 letters of S, x times.

Return the final string after all such shifts to S are applied.

Example 1:

Input: S = "abc", shifts = [3,5,9]
Output: "rpl"
Explanation: 
We start with "abc".
After shifting the first 1 letters of S by 3, we have "dbc".
After shifting the first 2 letters of S by 5, we have "igc".
After shifting the first 3 letters of S by 9, we have "rpl", the answer.

Note:

1. 1 <= S.length = shifts.length <= 20000
2. 0 <= shifts[i] <= 10 ^ 9

因为不知道把数字转成字符的函数,所以就开了两个对象来进行帮助转换，思路就是，把每个字符的总变换次数求出来，然后挨个求解

/**
 * @param {string} S
 * @param {number[]} shifts
 * @return {string}
 */
var shiftingLetters = function(S, shifts) {
  var tot2 = [];
  var fu ={
    0:'a',
    1:'b',
    2:'c',
    3:'d',
    4:'e',
    5:'f',
    6:'g',
    7:'h',
    8:'i',
    9:'j',
    10:'k',
    11:'l',
    12:'m',
    13:'n',
    14:'o',
    15:'p',
    16:'q',
    17:'r',
    18:'s',
    19:'t',
    20:'u',
    21:'v',
    22:'w',
    23:'x',
    24:'y',
    25:'z'
  }
  var index = {
    'a':0,
    'b':1,
    'c':2,
    'd':3,
    'e':4,
    'f':5,
    'g':6,
    'h':7,
    'i':8,
    'j':9,
    'k':10,
    'l':11,
    'm':12,
    'n':13,
    'o':14,
    'p':15,
    'q':16,
    'r':17,
    's':18,
    't':19,
    'u':20,
    'v':21,
    'w':22,
    'x':23,
    'y':24,
    'z':25
  }
  var len = shifts.length;
  var num = 0;
  var tot = [];
  for(let i = 0; i < len; i ++){
    if(i == 0)
      tot2[i] =  shifts[i];
    else tot2[i] = tot2[i - 1] + shifts[i];
    num += shifts[i];
  }
  for(let i = 0; i < len; i ++){
    if(i == 0) tot[i] = num;
    else tot[i] = num - tot2[i - 1];
  }
  console.log(tot);
  var len2 = S.length;
  var ans = '';
  for(let i = 0; i < len2; i ++){
    ans += fu[ (index[S[i] ] + tot[i] % 26) % 26 ];
    //console.log(ans);
  }
  return ans;
};