Commit 851f47f

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新增一些题解

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode.cn id=150 lang=cpp`
	`3`	`+ *`
	`4`	`+ * [150] 逆波兰表达式求值`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+class Solution {`
	`9`	`+public:`
	`10`	`+ int evalRPN(vector<string>& tokens) {`
	`11`	`+ stack<int> nums;`
	`12`	`+ for (auto& t : tokens) {`
	`13`	`+ if (t == "+" \|\| t == "-" \|\| t == "*" \|\| t == "/") {`
	`14`	`+ int a = nums.top();`
	`15`	`+ nums.pop();`
	`16`	`+ int b = nums.top();`
	`17`	`+ nums.pop();`
	`18`	`+ if (t == "+") {`
	`19`	`+ nums.push(b + a);`
	`20`	`+ }`
	`21`	`+ if (t == "-") {`
	`22`	`+ nums.push(b - a);`
	`23`	`+ }`
	`24`	`+ if (t == "*") {`
	`25`	`+ nums.push(b * a);`
	`26`	`+ }`
	`27`	`+ if (t == "/") {`
	`28`	`+ nums.push(b / a);`
	`29`	`+ }`
	`30`	`+ } else {`
	`31`	`+ nums.push(stoi(t));`
	`32`	`+ }`
	`33`	`+ }`
	`34`	`+ // printf("nums.size() = %d\n", nums.size());`
	`35`	`+ return nums.top();`
	`36`	`+ }`
	`37`	`+};`
	`38`	`+// @lc code=end`
	`39`	`+`

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode.cn id=152 lang=cpp`
	`3`	`+ *`
	`4`	`+ * [152] 乘积最大子数组`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+class Solution {`
	`9`	`+public:`
	`10`	`+ int maxProduct(vector<int>& nums) {`
	`11`	`+ int n = nums.size();`
	`12`	`+ int max = INT_MIN;`
	`13`	`+ int product = 1;`
	`14`	`+ // 区间一定出现在最左或者最右`
	`15`	`+ // 两边有 1. 两正或者两负, 可以扩大区间`
	`16`	`+ // 2. 一正一负, 也能扩大一边的区间`
	`17`	`+ for (int i = 0; i < n; i++) {`
	`18`	`+ product *= nums[i];`
	`19`	`+ max = product > max ? product : max;`
	`20`	`+ if (product == 0)`
	`21`	`+ product = 1;`
	`22`	`+ }`
	`23`	`+ product = 1;`
	`24`	`+ for (int i = n - 1; i >= 0; i--) {`
	`25`	`+ product *= nums[i];`
	`26`	`+ max = product > max ? product : max;`
	`27`	`+ if (product == 0)`
	`28`	`+ product = 1;`
	`29`	`+ }`
	`30`	`+ return max;`
	`31`	`+ }`
	`32`	`+};`
	`33`	`+// @lc code=end`
	`34`	`+`

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode.cn id=155 lang=cpp`
	`3`	`+ *`
	`4`	`+ * [155] 最小栈`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+class MinStack {`
	`9`	`+public:`
	`10`	`+ MinStack() {`
	`11`	`+`
	`12`	`+ }`
	`13`	`+`
	`14`	`+ void push(int val) {`
	`15`	`+ if (minv.empty()) {`
	`16`	`+ minv.push(val);`
	`17`	`+ } else {`
	`18`	`+ minv.push(min(val, minv.top()));`
	`19`	`+ }`
	`20`	`+ data.push(val);`
	`21`	`+ }`
	`22`	`+`
	`23`	`+ void pop() {`
	`24`	`+ data.pop();`
	`25`	`+ minv.pop();`
	`26`	`+ }`
	`27`	`+`
	`28`	`+ int top() {`
	`29`	`+ return data.top();`
	`30`	`+ }`
	`31`	`+`
	`32`	`+ int getMin() {`
	`33`	`+ return minv.top();`
	`34`	`+ }`
	`35`	`+`
	`36`	`+private:`
	`37`	`+ // 我们只需要设计一个数据结构,使得每个元素 a 与其相应的最小值 m 时刻保持一一对应`
	`38`	`+ stack<int> data;`
	`39`	`+ stack<int> minv;`
	`40`	`+};`
	`41`	`+`
	`42`	`+/**`
	`43`	`+ * Your MinStack object will be instantiated and called as such:`
	`44`	`+ * MinStack* obj = new MinStack();`
	`45`	`+ * obj->push(val);`
	`46`	`+ * obj->pop();`
	`47`	`+ * int param_3 = obj->top();`
	`48`	`+ * int param_4 = obj->getMin();`
	`49`	`+ */`
	`50`	`+// @lc code=end`
	`51`	`+`

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