Container With Most Water
2014-03-16 14:23
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题目原型:
Given n non-negative integers a1, a2,
..., an, where each represents a point at coordinate (i, ai). n vertical
lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together
with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.
基本思路:
首先,我们假设找到了最大的容积S,即由ai,aj和x轴围成,那么我们想一下,此时如果i的左边有比ai大的数,那么此时最大面积就不可能是S了(因为x轴的长度增加了,而高度最少是保证很原来相等的,当增加的是较长的边不影响高度,这是由短板理论决定的。),同理,j的右边也是如此。此时,我们得出了一个结论,当出现最大面积时,i的左边和j的右边的高度都比ai和aj小。
那么,现在考虑i和j的中间,当中间有比ai和aj大的数时,有可能出现比S更大的容积,因为尽管x轴减小了,当高度补齐了它的不足。所以,我们从两头向中间靠拢,同时更新候选值;在收缩区间的时候优先从 x, y中较小的边开始收缩(这点很重要,若是从较大的开始收缩会越来越小)。
Given n non-negative integers a1, a2,
..., an, where each represents a point at coordinate (i, ai). n vertical
lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together
with x-axis forms a container, such that the container contains the most water.
Note: You may not slant the container.
基本思路:
首先,我们假设找到了最大的容积S,即由ai,aj和x轴围成,那么我们想一下,此时如果i的左边有比ai大的数,那么此时最大面积就不可能是S了(因为x轴的长度增加了,而高度最少是保证很原来相等的,当增加的是较长的边不影响高度,这是由短板理论决定的。),同理,j的右边也是如此。此时,我们得出了一个结论,当出现最大面积时,i的左边和j的右边的高度都比ai和aj小。
那么,现在考虑i和j的中间,当中间有比ai和aj大的数时,有可能出现比S更大的容积,因为尽管x轴减小了,当高度补齐了它的不足。所以,我们从两头向中间靠拢,同时更新候选值;在收缩区间的时候优先从 x, y中较小的边开始收缩(这点很重要,若是从较大的开始收缩会越来越小)。
public int min(int i,int j) { return i<j?i:j; } public int maxArea(int[] height) { if(height==null||height.length==0) return 0; int area = 0; int maxArea = 0; for(int i = 0,j=height.length-1;i<j;) { area = min(height[i],height[j])*(j-i); maxArea = maxArea>area?maxArea:area; if(height[i]<height[j]) { int tmpi = i; while(tmpi<j&&height[tmpi]<=height[i]) { tmpi++; } i = tmpi; } else { int tmpj = j; while(tmpj>i&&height[tmpj]<=height[j]) { tmpj--; } j = tmpj; } } return maxArea; }
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