一开始觉得应该是个dp 题，把所有结果搜出来然后max 一下。实现以后发现组合太多了，非常慢，即使加上memorization 也是TLE

var hash = function(arr, p, s) {    arr = arr.sort();    return arr.reduce((p, c)=>p+"-"+c, "") + "_" + p + "_" + s;}let m = {};var iter = function(tokens, p, s) {    if (tokens.length == 0) return s;    let h = hash(tokens, p, s);    if (m[h] != void 0) return m[h];    let result = s;    for (let i = 0; i < tokens.length; ++i) {        //option1, if we have at leaset token[i] power        if (p >= tokens[i]) {            result = Math.max(result, iter(tokens.filter((x,k)=>k!=i), p-tokens[i], s+1));        }                //option2, if we have at least 1 point        if (s >= 1) {            result = Math.max(result, iter(tokens.filter((x,k)=>k!=i), p+tokens[i], s-1));        }    }    m[h] = Math.max(result, m[h]||0);    return result;}var bagOfTokensScore = function(tokens, P) {    return iter(tokens, P, 0);};

看了答案发现是用greedy，然而也没有证明为啥greedy 就是最优解。