494. Target Sum 添加标点符号求和
[抄题]:
You are given a list of non-negative integers, a1, a2, ..., an, and a target, S. Now you have 2 symbols +
and -
. For each integer, you should choose one from +
and -
as its new symbol.
Find out how many ways to assign symbols to make sum of integers equal to target S.
Example 1:
Input: nums is [1, 1, 1, 1, 1], S is 3. Output: 5 Explanation: -1+1+1+1+1 = 3 +1-1+1+1+1 = 3 +1+1-1+1+1 = 3 +1+1+1-1+1 = 3 +1+1+1+1-1 = 3 There are 5 ways to assign symbols to make the sum of nums be target 3.
[暴力解法]:
时间分析:
空间分析:
[优化后]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
种类:看似用DP,但是其实很麻烦
[一句话思路]:
用DFS也能求出种类
[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):
[画图]:
[一刷]:
- 理解一下退出条件:符合sum = target就count++,达到长度要求就退出
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
[复杂度]:Time complexity: O(每一个都试试加减法 2^n) Space complexity: O(1)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[算法思想:递归/分治/贪心]:递归
[关键模板化代码]:
数组名、目标、位置、当前和
public void dfs(int[] nums, int target, int pos, int sum) { //exit if (pos == nums.length) { if (sum == target) count++; return ; }
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
282. Expression Add Operators
[代码风格] :
class Solution { int count = 0; public int findTargetSumWays(int[] nums, int S) { //cc if (nums == null || nums.length == 0) return 0; //dfs dfs(nums, S, 0, 0); //return return count; } public void dfs(int[] nums, int target, int pos, int sum) { //exit if (pos == nums.length) { if (sum == target) count++; return ; } //dfs dfs(nums, target, pos + 1, sum + nums[pos]); dfs(nums, target, pos + 1, sum - nums[pos]); } }