标签：valueof different int for example 计算 rdl bsp ret
Given a string of numbers and operators,
return all possible results from computing all the different possible ways to group numbers and operators.
The valid operators are +, - and *. Example 1 Input: "2-1-1". ((2-1)-1) = 0 (2-(1-1)) = 2 Output: [0, 2] Example 2 Input: "2*3-4*5" (2*(3-(4*5))) = -34 ((2*3)-(4*5)) = -14 ((2*(3-4))*5) = -10 (2*((3-4)*5)) = -10 (((2*3)-4)*5) = 10 Output: [-34, -14, -10, -10, 10]
用到了Divide and Conquer, 跟 Leetcode: Unique Binary Search Trees II 很像
在input string里遍历各个operator, 依据每个operator分成左右子串，左右子串做递归返回所有可能的results，然后全排列。
注意很巧妙的一点在于寻找operator来划分，然后如果像20-23行那样没有划分成功（因为每找到operator）则Inter.parseInt(input)，这样省去了计算多位integer的工夫。比如input.charAt(i)=2, input.charAt(i+1)=3, 我们不需要再自己计算得到23.
dfs 调用自身即可,
分治法--后序遍历返回值, 先序遍历输入值. 画图自下而上返回.
注意 叶结点的处理
public class Solution {
public List<Integer> diffWaysToCompute(String input) {
List<Integer> res = new ArrayList<Integer>();
if (input==null || input.length()==0) return res;
for (int i=0; i<input.length(); i++) { //结束条件 i = input.length()
char c = input.charAt(i);
if (!isOperator(c)) continue; //递归条件
List<Integer> left = diffWaysToCompute(input.substring(0, i));
List<Integer> right = diffWaysToCompute(input.substring(i+1));
//递归条件---返回值
for (int j=0; j<left.size(); j++) {
for (int k=0; k<right.size(); k++) {
int a = left.get(j);
int b = right.get(k);
res.add(calc(a,b,c));
}
}
}
// 叶节点的处理
if (res.size() == 0) { //input is not null or size==0, but res.size()==0, meaning input is just a number
res.add(Integer.valueOf(input));
}
return res;
}
public boolean isOperator(char c) {
if (c==‘+‘ || c==‘-‘ || c==‘*‘) return true;
return false;
}
public int calc(int a, int b, char c) {
int res = Integer.MAX_VALUE;
switch(c) {
case ‘+‘: res = a+b; break;
case ‘-‘: res = a-b; break;
case ‘*‘: res = a*b; break;
}
return res;
}
}
Different Ways to Add Parentheses
标签：valueof different int for example 计算 rdl bsp ret
原文地址：http://www.cnblogs.com/apanda009/p/7296477.html
