++cnt1[s1.charAt(i) - ‘a‘]；

++cnt1[s1.charAt(i) - 'a']; among String s1

s1.charAt(i) Return string s1 Of the i Characters .

Suppose the string s1 Only lowercase letters （ For instance from 'a' To 'z'）,s1.charAt(i) - 'a' The returned character is a Of index 0,b  index 1, wait （z index25）

++cnt1[s1.charAt(i) - 'a'];// Match character s1.charAt(i) The corresponding counter plus 1.

therefore , This cycle for (int i = 0; i < n; ++i) {++cnt1[s1.charAt(i) - 'a']; } Compute string s1 The number of times each letter appears in .

Last , One cn1[0] Will include 'a' The number of , One cn1[1] Will include 'b' The number of , And so on .