【210607】494. 目标和

回溯

//官方题解
class Solution {
    int count = 0;

    public int findTargetSumWays(int[] nums, int target) {
        backtrack(nums, target, 0, 0);
        return count;
    }

    public void backtrack(int[] nums, int target, int index, int sum) {
        //出口
        if (index == nums.length) {
            if (sum == target) {
                count++;
            }
        } else {
        	//每个元素有两种添加符号的可能
            backtrack(nums, target, index + 1, sum + nums[index]);
            backtrack(nums, target, index + 1, sum - nums[index]);
        }
    }
}

动态规划

【0-1背包】子集划分

//官方题解
class Solution {
    public int findTargetSumWays(int[] nums, int target) {
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        int diff = sum - target;
        //出口：neg = (sum - target) / 2 是非负偶数
        if (diff < 0 || diff % 2 != 0) {
            return 0;
        }
        int neg = diff / 2;
        //根据背包的容量neg来动态规划
        //编程技巧：每一轮状态只与前一轮有关，于是省去一维
        int[] dp = new int[neg + 1];
        //不选择物品（容量为0）的方案有1个
        dp[0] = 1;
        //循环每个物品为一维
        for (int num : nums) {
            for (int j = neg; j >= num; j--) {
                dp[j] += dp[j - num];
            }
        }
        return dp[neg];
    }
}