problem:
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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.
X轴为底,两个纵轴为变,求容器的容积,短边是瓶颈。
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thinking:
(1)短边决定水箱的有效高,底要尽可能的宽。
(2)典型的双指针求解的题型。
(3)贪心的策略,哪条边短,往里收缩寻找下一条边。
code:
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class Solution {
public:
int maxArea(vector
&height) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int i = 0;
int j = height.size() - 1;
int ret = 0;
while(i < j)
{
int area = (j - i) * min(height[i], height[j]);
ret = max(ret, area);
if (height[i] <= height[j])
i++;
else
j--;
}
return ret;
}
};
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时间复杂度为O(n)
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暴力 破解法:时间复杂度为O(n*n)
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int area(vector::iterator &a,vector ::iterator &b) { return (b-a)*(*a>*b?*b:*a); } class Solution { public: int maxArea(vector &height) { int max_area=0; for(vector ::iterator i=height.begin()+1;i!=height.end();i++) { for(vector ::iterator j=height.begin();j!=i;j++) { max_area=max(max_area,area(j,i)); } } return max_area; } };
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