[LeetCode] 494. Target Sum（目标和）

• Difficulty: Medium

• Related Topics: Dynamic Programming, Depth-first Search

• Link: https://leetcode.com/problems/target-sum/

Description

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.
给定一个包含非负整数的列表 [a1, a2, ..., an]，以及一个目标 S。你有两个符号 + 和 -，对于每一个整数，你需要从这两个符号中选择一个作为这个数的符号。

Find out how many ways to assign symbols to make sum of integers equal to target S.
找出有多少种赋值的方式，使得赋值过后这些数的和等于 S。

Example

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.

Constraints

• The length of the given array is positive and will not exceed 20.
  给定的数组保证非空，且长度不超过 20。
• The sum of elements in the given array will not exceed 1000.
  给定数组的元素和不超过 1000。
• Your output answer is guaranteed to be fitted in a 32-bit integer.
  你的答案保证在 32 位整数的范围内。

Solution

由于此题只出现 + 和 -，没有其它符号，所以简单的深搜即可完成任务。代码如下：

class Solution {
    private var result = 0

    fun findTargetSumWays(nums: IntArray, S: Int): Int {
        backtrack(nums, S, 0, 0)
        return result
    }

    private fun backtrack(nums: IntArray, s: Int, curSum: Int, curIndex: Int) {
        if (curIndex == nums.size ) {
            if (curSum == s) {
                result++
            }
            return
        }
        // 当前数的符号取 +
        backtrack(nums, s, curSum + nums[curIndex], curIndex + 1)
        // 当前数的符号取 -
        backtrack(nums, s, curSum - nums[curIndex], curIndex + 1)
    }
}