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`3`	`3`	对于区间的问题,我们会思考用区间型的DP是否可行。我们自然会令dp[i]表示针对前i个元素,变换成功所需要的最少操作次数。接下来就是要考虑最后一个区间的位置起点j。因为本题里字符串的长度只有100,我们可以考虑遍历起点j。这样问题就转化为对于区间[j:i],变化成功所需要的最少操作次数,这样就有`dp[i]=dp[j-1]+helper(j,i)`.
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