java 解法, 执行用时: 1 ms, 内存消耗: 39.7 MB, 提交时间: 2023-06-08 10:50:01

class Solution {
    public List<Integer> splitIntoFibonacci(String S) {
        List<Integer> res = new ArrayList<>();
        backtrack(S.toCharArray(), res, 0);
        return res;
    }

    public boolean backtrack(char[] digit, List<Integer> res, int index) {
        //边界条件判断，如果截取完了，并且res长度大于等于3，表示找到了一个组合。
        if (index == digit.length && res.size() >= 3) {
            return true;
        }
        for (int i = index; i < digit.length; i++) {
            //两位以上的数字不能以0开头
            if (digit[index] == '0' && i > index) {
                break;
            }
            //截取字符串转化为数字
            long num = subDigit(digit, index, i + 1);
            //如果截取的数字大于int的最大值，则终止截取
            if (num > Integer.MAX_VALUE) {
                break;
            }
            int size = res.size();
            //如果截取的数字大于res中前两个数字的和，说明这次截取的太大，直接终止，因为后面越截取越大
            if (size >= 2 && num > res.get(size - 1) + res.get(size - 2)) {
                break;
            }
            if (size <= 1 || num == res.get(size - 1) + res.get(size - 2)) {
                //把数字num添加到集合res中
                res.add((int) num);
                //如果找到了就直接返回
                if (backtrack(digit, res, i + 1))
                    return true;
                //如果没找到，就会走回溯这一步，然后把上一步添加到集合res中的数字给移除掉
                res.remove(res.size() - 1);
            }
        }
        return false;
    }

    //相当于截取字符串S中的子串然后转换为十进制数字
    private long subDigit(char[] digit, int start, int end) {
        long res = 0;
        for (int i = start; i < end; i++) {
            res = res * 10 + digit[i] - '0';
        }
        return res;
    }
}

java 解法, 执行用时: 0 ms, 内存消耗: 39.8 MB, 提交时间: 2023-06-08 10:49:20

class Solution {
    public List<Integer> splitIntoFibonacci(String num) {
        List<Integer> list = new ArrayList<Integer>();
        backtrack(list, num, num.length(), 0, 0, 0);
        return list;
    }

    public boolean backtrack(List<Integer> list, String num, int length, int index, int sum, int prev) {
        if (index == length) {
            return list.size() >= 3;
        }
        long currLong = 0;
        for (int i = index; i < length; i++) {
            if (i > index && num.charAt(index) == '0') {
                break;
            }
            currLong = currLong * 10 + num.charAt(i) - '0';
            if (currLong > Integer.MAX_VALUE) {
                break;
            }
            int curr = (int) currLong;
            if (list.size() >= 2) {
                if (curr < sum) {
                    continue;
                } else if (curr > sum) {
                    break;
                }
            }
            list.add(curr);
            if (backtrack(list, num, length, i + 1, prev + curr, curr)) {
                return true;
            } else {
                list.remove(list.size() - 1);
            }
        }
        return false;
    }
}

javascript 解法, 执行用时: 80 ms, 内存消耗: 42.4 MB, 提交时间: 2023-06-08 10:49:04

/**
 * @param {string} num
 * @return {number[]}
 */
var splitIntoFibonacci = function(num) {
    const list = new Array().fill(0);
    backtrack(list, num, num.length, 0, 0, 0);
    return list;
};

const backtrack = (list, num, length, index, sum, prev) => {
    if (index === length) {
        return list.length >= 3;
    }
    let currLong = 0;
    for (let i = index; i < length; i++) {
        if (i > index && num[index] === '0') {
            break;
        }
        currLong = currLong * 10 + num[i].charCodeAt() - '0'.charCodeAt();
        if (currLong > Math.pow(2, 31) - 1) {
            break;
        }
        let curr = currLong;
        if (list.length >= 2) {
            if (curr < sum) {
                continue;
            } else if (curr > sum) {
                break;
            }
        }
        list.push(curr);
        if (backtrack(list, num, length, i + 1, prev + curr, curr)) {
            return true;
        } else {
            list.splice(list.length - 1, 1);
        }
    }
    return false;
}

golang 解法, 执行用时: 4 ms, 内存消耗: 1.9 MB, 提交时间: 2023-06-08 10:48:48

func splitIntoFibonacci(num string) (F []int) {
    n := len(num)
    var backtrack func(index, sum, prev int) bool
    backtrack = func(index, sum, prev int) bool {
        if index == n {
            return len(F) >= 3
        }

        cur := 0
        for i := index; i < n; i++ {
            // 每个块的数字一定不要以零开头，除非这个块是数字 0 本身
            if i > index && num[index] == '0' {
                break
            }

            cur = cur*10 + int(num[i]-'0')
            // 拆出的整数要符合 32 位有符号整数类型
            if cur > math.MaxInt32 {
                break
            }

            // F[i] + F[i+1] = F[i+2]
            if len(F) >= 2 {
                if cur < sum {
                    continue
                }
                if cur > sum {
                    break
                }
            }

            // cur 符合要求，加入序列 F
            F = append(F, cur)
            if backtrack(i+1, prev+cur, cur) {
                return true
            }
            F = F[:len(F)-1]
        }
        return false
    }
    backtrack(0, 0, 0)
    return
}

python3 解法, 执行用时: 60 ms, 内存消耗: 16 MB, 提交时间: 2023-06-08 10:48:34

# 回溯 + 剪枝
class Solution:
    def splitIntoFibonacci(self, num: str) -> List[int]:
        ans = list()

        def backtrack(index: int):
            if index == len(num):
                return len(ans) >= 3
            
            curr = 0
            for i in range(index, len(num)):
                if i > index and num[index] == "0":
                    break
                curr = curr * 10 + ord(num[i]) - ord("0")
                if curr > 2**31 - 1:
                    break
                
                if len(ans) < 2 or curr == ans[-2] + ans[-1]:
                    ans.append(curr)
                    if backtrack(i + 1):
                        return True
                    ans.pop()
                elif len(ans) > 2 and curr > ans[-2] + ans[-1]:
                    break
        
            return False
        
        backtrack(0)
        return ans