https://leetcode.com/problems/bag-of-tokens/
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。