greedy/container-with-most-water
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2023-12-01
Container With Most Water
描述
Given n
non-negative integers $$a_1, a_2, ..., a_n$$, where each represents a point at coordinate $$(i, a_i)$$. n vertical lines are drawn such that the two endpoints of line i
is at $$(i, a_i)$$ 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.
分析
每个容器的面积,取决于最短的木板。
代码
// Container With Most Water
// 时间复杂度O(n),空间复杂度O(1)
public class Solution {
public int maxArea(int[] height) {
int start = 0;
int end = height.length - 1;
int result = Integer.MIN_VALUE;
while (start < end) {
int area = Math.min(height[end], height[start]) * (end - start);
result = Math.max(result, area);
if (height[start] <= height[end]) {
start++;
} else {
end--;
}
}
return result;
}
}
// LeetCode, Container With Most Water
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
int maxArea(vector<int> &height) {
int start = 0;
int end = height.size() - 1;
int result = INT_MIN;
while (start < end) {
int area = min(height[end], height[start]) * (end - start);
result = max(result, area);
if (height[start] <= height[end]) {
start++;
} else {
end--;
}
}
return result;
}
};