81 lines
1.8 KiB
Java
81 lines
1.8 KiB
Java
//请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都
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//是O(1)。
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//
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// 若队列为空，pop_front 和 max_value 需要返回 -1
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//
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// 示例 1：
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//
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// 输入:
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//["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]
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//[[],[1],[2],[],[],[]]
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//输出: [null,null,null,2,1,2]
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//
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//
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// 示例 2：
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//
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// 输入:
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//["MaxQueue","pop_front","max_value"]
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//[[],[],[]]
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//输出: [null,-1,-1]
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//
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//
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//
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//
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// 限制：
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//
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//
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// 1 <= push_back,pop_front,max_value的总操作数 <= 10000
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// 1 <= value <= 10^5
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//
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// Related Topics 栈 Sliding Window
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// 👍 248 👎 0
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package leetcode.editor.cn;
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//剑指 Offer 59 - II:队列的最大值
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public class DuiLieDeZuiDaZhiLcof {
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public static void main(String[] args) {
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//测试代码
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// Solution solution = new DuiLieDeZuiDaZhiLcof().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class MaxQueue {
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int[] arr = new int[20000];
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int start = 0, end = 0;
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public MaxQueue() {
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}
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public int max_value() {
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int ans = -1;
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for (int i = start; i != end; ++i) {
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ans = Math.max(ans, arr[i]);
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}
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return ans;
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}
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public void push_back(int value) {
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arr[end++] = value;
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}
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public int pop_front() {
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if (start == end) {
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return -1;
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}
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return arr[start++];
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}
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}
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/**
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* Your MaxQueue object will be instantiated and called as such:
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* MaxQueue obj = new MaxQueue();
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* int param_1 = obj.max_value();
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* obj.push_back(value);
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* int param_3 = obj.pop_front();
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*/
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//leetcode submit region end(Prohibit modification and deletion)
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} |