LeetCode 11 Container With Most Water
2017-01-09 15:25
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题意:
数组A中第i个元素A[i]表示一块两端在(i,0)和(i,A[i])的板,从中选出2块板,使得它们与x轴围成的桶装水体积最大。
思路:
因为装水体积只和矮的板有关,因此我们从高到低的考虑板子,这样当前考虑的板子A[i]就是装水的高度。
高度确定时,让桶尽量的宽。可以用l和r表示已经考虑过的板子的x坐标的最小和最大值。
若l <= i <= r,则桶宽max( i - l, r - i );若i < l,则桶宽r - i;若i > r,则桶宽i - l。
时间复杂度O(nlogn)。
官方题解中的O(n)做法是错误的,解法没有覆盖所有的可能,有点类似“用贪心策略做dp而漏解”。
代码:
数组A中第i个元素A[i]表示一块两端在(i,0)和(i,A[i])的板,从中选出2块板,使得它们与x轴围成的桶装水体积最大。
思路:
因为装水体积只和矮的板有关,因此我们从高到低的考虑板子,这样当前考虑的板子A[i]就是装水的高度。
高度确定时,让桶尽量的宽。可以用l和r表示已经考虑过的板子的x坐标的最小和最大值。
若l <= i <= r,则桶宽max( i - l, r - i );若i < l,则桶宽r - i;若i > r,则桶宽i - l。
时间复杂度O(nlogn)。
官方题解中的O(n)做法是错误的,解法没有覆盖所有的可能,有点类似“用贪心策略做dp而漏解”。
代码:
// // Created by house on 1/9/17. // class Solution { public: int maxArea(vector<int> &height) { int ans = 0; vector <pair<int, int>> rank; for (int i = 0; i < height.size(); ++i) { rank.push_back(make_pair(height[i], i)); } sort(rank.begin(), rank.end()); int l, r; l = r = rank.rbegin()->second; for (int i = rank.size() - 2; i >= 0; --i) { int pos = rank[i].second, w = 0; if (pos < l) { w = r - pos; l = pos; } else if (pos > r) { w = pos - l; r = pos; } else { w = max(r - pos, pos - l); } if (ans < w * rank[i].first) { ans = w * rank[i].first; } } return ans; } };
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