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	`1`	`+# No.53 最大子序和`
	`2`	`+`
	`3`	+难度:`easy`
	`4`	`+`
	`5`	`+给定一个整数数组 nums ,找到一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和。`
	`6`	`+`
	`7`	`+## 示例`
	`8`	`+`
	`9`	`+`
	`10`	`+`
	`11`	`+示例:`
	`12`	+```
	`13`	`+输入: [-2,1,-3,4,-1,2,1,-5,4],`
	`14`	`+输出: 6`
	`15`	`+解释: 连续子数组 [4,-1,2,1] 的和最大,为 6。`
	`16`	+```
	`17`	`+`
	`18`	`+## 进阶`
	`19`	`+`
	`20`	`+进阶:如果你已经实现复杂度为 O(n) 的解法,尝试使用更为精妙的分治法求解。`
	`21`	`+`
	`22`	`+## 解题思路`
	`23`	`+`
	`24`	`+思路一:`
	`25`	`+`
	`26`	`+暴力法,双重循环,每次遍历求和并保留最大值。`
	`27`	`+`
	`28`	`+代码如下:`
	`29`	`+`
	`30`	+```javascript
	`31`	`+/**`
	`32`	`+ * @param {number[]} nums`
	`33`	`+ * @return {number}`
	`34`	`+ */`
	`35`	`+var maxSubArray = function(nums) {`
	`36`	`+ //暴力法`
	`37`	`+ let max = -Infinity;`
	`38`	`+ for (let i = 0, n = nums.length; i < n; i++) {`
	`39`	`+ let total = 0;`
	`40`	`+ for (let j = i; j < n; j++) {`
	`41`	`+ total += nums[j];`
	`42`	`+ if (total > max)`
	`43`	`+ max = total;`
	`44`	`+ }`
	`45`	`+ }`
	`46`	`+ return max;`
	`47`	`+};`
	`48`	+```
	`49`	`+`
	`50`	`+思路二:`
	`51`	`+`
	`52`	`+动态规划,一遍遍历,每次遍历计算出。类似寻找最大最小值的题目,初始值一定要定义成理论上的最小最大值`

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