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add golang solution 279

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode.cn id=279 lang=golang`
	`3`	`+ * 动态规划的思路,状态转移方程:dp[n] = min(dp[n-11]+1, dp[n-22]+1, ..., dp[n-kk]+1), ( 0< kk <=n )`
	`4`	`+ */`
	`5`	`+func numSquares(n int) int {`
	`6`	`+ if n <= 0 {`
	`7`	`+ return 0`
	`8`	`+ }`
	`9`	`+ dp := make([]int, n+1) // 多申请了一份整形,使代码更容易理解, dp[n] 就是 n 的完全平方数的求解`
	`10`	`+ for i := 1; i <= n; i++ {`
	`11`	`+ dp[i] = i // 初始值 dp[n] 的最大值的解,也是最容易求的解`
	`12`	`+ for j := 0; j*j <= i; j++ {`
	`13`	`+ dp[i] = minInt(dp[i-j*j]+1, dp[i])`
	`14`	`+ }`
	`15`	`+ }`
	`16`	`+ return dp[n]`
	`17`	`+}`
	`18`	`+`
	`19`	`+func minInt(x, y int) int {`
	`20`	`+ if x < y {`
	`21`	`+ return x`
	`22`	`+ }`
	`23`	`+ return y`
	`24`	`+}`

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