Leetcode 842 将数组拆分成斐波那契序列

 

　　回溯解法 JAVA：

    public final List<Integer> splitIntoFibonacci(String S) {
        List<Integer> reList = new LinkedList<Integer>();
        search(S, 0, reList, 0);
        return reList;
    }

    private final boolean search(String s, int flag, List<Integer> reList, int reLen) {
        if (flag == s.length() && reLen > 2) {
            return true;
        }
        int nextLen = reLen + 1;
        for (int i = flag + 1; i <= s.length(); i++) {
            String inStr = s.substring(flag, i);
            if (i > flag + 1 && '0' == inStr.charAt(0)) {
                continue;
            }
            if (i > flag + 10) {
                break;
            }
            long numLon = Long.valueOf(inStr);
            if (numLon > Integer.MAX_VALUE) {
                continue;
            }
            if (reLen >= 2) {
                if (reList.get(reLen - 2) + reList.get(reLen - 1) != numLon) {
                    continue;
                }
            }
            reList.add((int) numLon);
            if (search(s, i, reList, nextLen)) {
                return true;
            }
            reList.remove(reList.size() - 1);
        }
        return false;
    }

　　回溯解法 JS：

var splitIntoFibonacci = function(S) {
    let reArr = [];
    search(S,reArr,0);
    return reArr;
};

var search=function(s,reArr,flag){
    if(flag==s.length){
        if(reArr.length>2){
            return true;
        }
        return false;
    }
    for(let i=flag+1;i<=s.length;i++){
        let current = s.slice(flag,i);
        let arrLen = reArr.length;
        if(i-flag>1&&current.charAt(0)=='0'){
            continue;
        }
        if(arrLen>=2){
            if(Number(current)>2**31-1){continue;}
            let sum = Number(reArr[arrLen-1])+Number(reArr[arrLen-2]);
            if(sum!=Number(current)){
                continue;
            }
        }
        reArr.push(current);
        if(search(s,reArr,i)){
            return true;
        }
        reArr.pop();
    }
    return false;
}