Sunday, November 25, 2018
948. Bag of Tokens
Version #1 Greedy
Buy at the cheapest and sell at the most expensive
Time O(nlogn)
class Solution {
public int bagOfTokensScore(int[] tokens, int P) {
Arrays.sort(tokens);
int left = 0;
int point = 0;
int max = 0;
for (int right = tokens.length - 1; right >= left; right--) {
while (left <= right && P >= tokens[left]) {
P -= tokens[left++];
point++;
}
max = Math.max(max, point);
if (point > 0) {
P += tokens[right];
point--;
}
}
return max;
}
}
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