Commit 051dfbf

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20190312

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	`1`	`+package code;`
	`2`	`+`
	`3`	`+import java.util.List;`
	`4`	`+/*`
	`5`	`+ * 120. Triangle`
	`6`	`+ * 题意:三角形的矩阵,求值最小的路径`
	`7`	`+ * 难度:Medium`
	`8`	`+ * 分类:Array, Dynamic Porgramming`
	`9`	`+ * 思路:动态规划 dp[i] = min(dp[i-1],dp[i]) + val`
	`10`	`+ * Tips:`
	`11`	`+ */`
	`12`	`+public class lc120 {`
	`13`	`+ public int minimumTotal(List<List<Integer>> triangle) {`
	`14`	`+ int len = triangle.get(triangle.size()-1).size();`
	`15`	`+ int[] dp = new int[len];`
	`16`	`+ int[] dp2 = new int[len];`
	`17`	`+ int index = 0;`
	`18`	`+ int res = Integer.MAX_VALUE;`
	`19`	`+ while(index<triangle.size()){`
	`20`	`+ List<Integer> ls = triangle.get(index);`
	`21`	`+ res = Integer.MAX_VALUE;`
	`22`	`+ for (int i = 0; i < ls.size() ; i++) {`
	`23`	`+ if(i==0) dp2[i] = dp[i] + ls.get(i);`
	`24`	`+ else if(i==ls.size()-1) dp2[i] = dp[i-1] + ls.get(i); //注意是 i==ls.size()-1`
	`25`	`+ else dp2[i] = Math.min(dp[i-1],dp[i]) + ls.get(i);`
	`26`	`+ res = Math.min(dp2[i], res);`
	`27`	`+ }`
	`28`	`+ dp = dp2; //两个一维数组`
	`29`	`+ dp2 = new int[len];`
	`30`	`+ index++;`
	`31`	`+ }`
	`32`	`+ return res;`
	`33`	`+ }`
	`34`	`+}`

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	`1`	`+package code;`
	`2`	`+`
	`3`	`+import java.util.Arrays;`
	`4`	`+/*`
	`5`	`+ * 213. House Robber II`
	`6`	`+ * 题意:数组最大和,不能选取相邻的两个数。数组首尾是连着的`
	`7`	`+ * 难度:Medium`
	`8`	`+ * 分类:Dynamic Programming`
	`9`	`+ * 思路:分别把第一个元素置0和最后一个元素置0,用lc198的解法`
	`10`	`+ *`
	`11`	`+ * Tips:lc198`
	`12`	`+ */`
	`13`	`+public class lc213 {`
	`14`	`+ public int rob(int[] nums) { //分别把第一个元素置0和最后一个元素置0,用lc198的解法`
	`15`	`+ if(nums.length == 0)`
	`16`	`+ return 0;`
	`17`	`+ if(nums.length == 1)`
	`18`	`+ return nums[0];`
	`19`	`+ int res1 = helper(Arrays.copyOf(nums, nums.length-1));`
	`20`	`+ nums[0] = 0;`
	`21`	`+ int res2 = helper(nums);`
	`22`	`+ return Math.max(res1, res2);`
	`23`	`+ }`
	`24`	`+`
	`25`	`+ public int helper(int[] nums) {`
	`26`	`+ if(nums.length == 0)`
	`27`	`+ return 0;`
	`28`	`+ if(nums.length == 1)`
	`29`	`+ return nums[0];`
	`30`	`+ int[] dp = new int[nums.length];`
	`31`	`+ dp[0] = nums[0];`
	`32`	`+ dp[1] = Math.max(nums[0], nums[1]);`
	`33`	`+ for (int i = 2; i < nums.length ; i++) {`
	`34`	`+ dp[i] = Math.max((dp[i-2] + nums[i]),dp[i-1]); //dp[i] 表示以 0~i 的数组的结果`
	`35`	`+ }`
	`36`	`+ return dp[nums.length-1];`
	`37`	`+ }`
	`38`	`+}`

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